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I need an example on how to paramtrize Vector of registers in terms of bit-width and initial values which are not '0' and are different for each register.
My use-case is a generic filter coefficients bank with some unique reset values to each, and off course an option to override values.
I thought of something like the below code (not really sure how to write the iteration, so this is kind of pseudo):
class Coeffbank(bitWidth : UInt ,ncoeff : UInt, rstVal : Vec(SInt)) extends Module {
// how do iterate through the reset vector ?? //
val coeffs = Vec.fill(ncoeff) {Reg(init = SInt(rstVal(i),width = bitwidth))
}
Also, when new'ing the above (instantiating this module how do I pass the list of reset value in the argument list?
Hoping to get some help on how to write it properly.
The explanation should probably be a bit more thorough, but basically you need to create a Reg of Vec. Something like should do it:
val coeffs = RegInit(rstVal)
In this case, since you already have the Vec of reset values, you can just pass it to the Reg constructor.
I'm assuming that the size of rstVal is equal to ncoeff, otherwise you'll need to reduce the size of rstVal with something like rstVal.take(ncoeff). Also note that I'm using RegInit which is the preferred way to create a register with a reset value.
Let's start with the easy case. This would be much easier if instead of a Vec of SInts your rstVal array was instead a scala collection (Seq, Array, ...) of regular SInt. When possible it is best to save generation of actual hardware until you directly need them. If rstVal contains Int's. Your code would become
val newRstVals = VecInit(Seq.tabulate(ncoeff) { index => rstVals(index).S(bitWidth.W) })
val reg = RegInit(newRstVals)
If you really need to pass in a Vec then the right approach is to create a separate type instance and use the two argument call to RegInit
val vecType = Vec(ncoeff, SInt(bitWidth.W))
val newRstVals1 = VecInit(Seq.tabulate(ncoeff) { index => newRstVals(index) })
val reg = RegInit(vecType, newRstVals1)
There might be problems if the bitWidth you pass in is not big enough to contain the constants you have passed in. You probably should have some checks for that.
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I've been using the == operator in my program to compare all my strings so far.
However, I ran into a bug, changed one of them into .equals() instead, and it fixed the bug.
Is == bad? When should it and should it not be used? What's the difference?
== tests for reference equality (whether they are the same object).
.equals() tests for value equality (whether they contain the same data).
Objects.equals() checks for null before calling .equals() so you don't have to (available as of JDK7, also available in Guava).
Consequently, if you want to test whether two strings have the same value you will probably want to use Objects.equals().
// These two have the same value
new String("test").equals("test") // --> true
// ... but they are not the same object
new String("test") == "test" // --> false
// ... neither are these
new String("test") == new String("test") // --> false
// ... but these are because literals are interned by
// the compiler and thus refer to the same object
"test" == "test" // --> true
// ... string literals are concatenated by the compiler
// and the results are interned.
"test" == "te" + "st" // --> true
// ... but you should really just call Objects.equals()
Objects.equals("test", new String("test")) // --> true
Objects.equals(null, "test") // --> false
Objects.equals(null, null) // --> true
You almost always want to use Objects.equals(). In the rare situation where you know you're dealing with interned strings, you can use ==.
From JLS 3.10.5. String Literals:
Moreover, a string literal always refers to the same instance of class String. This is because string literals - or, more generally, strings that are the values of constant expressions (§15.28) - are "interned" so as to share unique instances, using the method String.intern.
Similar examples can also be found in JLS 3.10.5-1.
Other Methods To Consider
String.equalsIgnoreCase() value equality that ignores case. Beware, however, that this method can have unexpected results in various locale-related cases, see this question.
String.contentEquals() compares the content of the String with the content of any CharSequence (available since Java 1.5). Saves you from having to turn your StringBuffer, etc into a String before doing the equality comparison, but leaves the null checking to you.
== tests object references, .equals() tests the string values.
Sometimes it looks as if == compares values, because Java does some behind-the-scenes stuff to make sure identical in-line strings are actually the same object.
For example:
String fooString1 = new String("foo");
String fooString2 = new String("foo");
// Evaluates to false
fooString1 == fooString2;
// Evaluates to true
fooString1.equals(fooString2);
// Evaluates to true, because Java uses the same object
"bar" == "bar";
But beware of nulls!
== handles null strings fine, but calling .equals() from a null string will cause an exception:
String nullString1 = null;
String nullString2 = null;
// Evaluates to true
System.out.print(nullString1 == nullString2);
// Throws a NullPointerException
System.out.print(nullString1.equals(nullString2));
So if you know that fooString1 may be null, tell the reader that by writing
System.out.print(fooString1 != null && fooString1.equals("bar"));
The following are shorter, but it’s less obvious that it checks for null:
System.out.print("bar".equals(fooString1)); // "bar" is never null
System.out.print(Objects.equals(fooString1, "bar")); // Java 7 required
== compares Object references.
.equals() compares String values.
Sometimes == gives illusions of comparing String values, as in following cases:
String a="Test";
String b="Test";
if(a==b) ===> true
This is because when you create any String literal, the JVM first searches for that literal in the String pool, and if it finds a match, that same reference will be given to the new String. Because of this, we get:
(a==b) ===> true
String Pool
b -----------------> "test" <-----------------a
However, == fails in the following case:
String a="test";
String b=new String("test");
if (a==b) ===> false
In this case for new String("test") the statement new String will be created on the heap, and that reference will be given to b, so b will be given a reference on the heap, not in String pool.
Now a is pointing to a String in the String pool while b is pointing to a String on the heap. Because of that we get:
if(a==b) ===> false.
String Pool
"test" <-------------------- a
Heap
"test" <-------------------- b
While .equals() always compares a value of String so it gives true in both cases:
String a="Test";
String b="Test";
if(a.equals(b)) ===> true
String a="test";
String b=new String("test");
if(a.equals(b)) ===> true
So using .equals() is always better.
The == operator checks to see if the two strings are exactly the same object.
The .equals() method will check if the two strings have the same value.
Strings in Java are immutable. That means whenever you try to change/modify the string you get a new instance. You cannot change the original string. This has been done so that these string instances can be cached. A typical program contains a lot of string references and caching these instances can decrease the memory footprint and increase the performance of the program.
When using == operator for string comparison you are not comparing the contents of the string, but are actually comparing the memory address. If they are both equal it will return true and false otherwise. Whereas equals in string compares the string contents.
So the question is if all the strings are cached in the system, how come == returns false whereas equals return true? Well, this is possible. If you make a new string like String str = new String("Testing") you end up creating a new string in the cache even if the cache already contains a string having the same content. In short "MyString" == new String("MyString") will always return false.
Java also talks about the function intern() that can be used on a string to make it part of the cache so "MyString" == new String("MyString").intern() will return true.
Note: == operator is much faster than equals just because you are comparing two memory addresses, but you need to be sure that the code isn't creating new String instances in the code. Otherwise you will encounter bugs.
String a = new String("foo");
String b = new String("foo");
System.out.println(a == b); // prints false
System.out.println(a.equals(b)); // prints true
Make sure you understand why. It's because the == comparison only compares references; the equals() method does a character-by-character comparison of the contents.
When you call new for a and b, each one gets a new reference that points to the "foo" in the string table. The references are different, but the content is the same.
Yea, it's bad...
== means that your two string references are exactly the same object. You may have heard that this is the case because Java keeps sort of a literal table (which it does), but that is not always the case. Some strings are loaded in different ways, constructed from other strings, etc., so you must never assume that two identical strings are stored in the same location.
Equals does the real comparison for you.
Yes, == is bad for comparing Strings (any objects really, unless you know they're canonical). == just compares object references. .equals() tests for equality. For Strings, often they'll be the same but as you've discovered, that's not guaranteed always.
Java have a String pool under which Java manages the memory allocation for the String objects. See String Pools in Java
When you check (compare) two objects using the == operator it compares the address equality into the string-pool. If the two String objects have the same address references then it returns true, otherwise false. But if you want to compare the contents of two String objects then you must override the equals method.
equals is actually the method of the Object class, but it is Overridden into the String class and a new definition is given which compares the contents of object.
Example:
stringObjectOne.equals(stringObjectTwo);
But mind it respects the case of String. If you want case insensitive compare then you must go for the equalsIgnoreCase method of the String class.
Let's See:
String one = "HELLO";
String two = "HELLO";
String three = new String("HELLO");
String four = "hello";
one == two; // TRUE
one == three; // FALSE
one == four; // FALSE
one.equals(two); // TRUE
one.equals(three); // TRUE
one.equals(four); // FALSE
one.equalsIgnoreCase(four); // TRUE
I agree with the answer from zacherates.
But what you can do is to call intern() on your non-literal strings.
From zacherates example:
// ... but they are not the same object
new String("test") == "test" ==> false
If you intern the non-literal String equality is true:
new String("test").intern() == "test" ==> true
== compares object references in Java, and that is no exception for String objects.
For comparing the actual contents of objects (including String), one must use the equals method.
If a comparison of two String objects using == turns out to be true, that is because the String objects were interned, and the Java Virtual Machine is having multiple references point to the same instance of String. One should not expect that comparing one String object containing the same contents as another String object using == to evaluate as true.
.equals() compares the data in a class (assuming the function is implemented).
== compares pointer locations (location of the object in memory).
== returns true if both objects (NOT TALKING ABOUT PRIMITIVES) point to the SAME object instance.
.equals() returns true if the two objects contain the same data equals() Versus == in Java
That may help you.
== performs a reference equality check, whether the 2 objects (strings in this case) refer to the same object in the memory.
The equals() method will check whether the contents or the states of 2 objects are the same.
Obviously == is faster, but will (might) give false results in many cases if you just want to tell if 2 Strings hold the same text.
Definitely the use of the equals() method is recommended.
Don't worry about the performance. Some things to encourage using String.equals():
Implementation of String.equals() first checks for reference equality (using ==), and if the 2 strings are the same by reference, no further calculation is performed!
If the 2 string references are not the same, String.equals() will next check the lengths of the strings. This is also a fast operation because the String class stores the length of the string, no need to count the characters or code points. If the lengths differ, no further check is performed, we know they cannot be equal.
Only if we got this far will the contents of the 2 strings be actually compared, and this will be a short-hand comparison: not all the characters will be compared, if we find a mismatching character (at the same position in the 2 strings), no further characters will be checked.
When all is said and done, even if we have a guarantee that the strings are interns, using the equals() method is still not that overhead that one might think, definitely the recommended way. If you want an efficient reference check, then use enums where it is guaranteed by the language specification and implementation that the same enum value will be the same object (by reference).
If you're like me, when I first started using Java, I wanted to use the "==" operator to test whether two String instances were equal, but for better or worse, that's not the correct way to do it in Java.
In this tutorial I'll demonstrate several different ways to correctly compare Java strings, starting with the approach I use most of the time. At the end of this Java String comparison tutorial I'll also discuss why the "==" operator doesn't work when comparing Java strings.
Option 1: Java String comparison with the equals method
Most of the time (maybe 95% of the time) I compare strings with the equals method of the Java String class, like this:
if (string1.equals(string2))
This String equals method looks at the two Java strings, and if they contain the exact same string of characters, they are considered equal.
Taking a look at a quick String comparison example with the equals method, if the following test were run, the two strings would not be considered equal because the characters are not the exactly the same (the case of the characters is different):
String string1 = "foo";
String string2 = "FOO";
if (string1.equals(string2))
{
// this line will not print because the
// java string equals method returns false:
System.out.println("The two strings are the same.")
}
But, when the two strings contain the exact same string of characters, the equals method will return true, as in this example:
String string1 = "foo";
String string2 = "foo";
// test for equality with the java string equals method
if (string1.equals(string2))
{
// this line WILL print
System.out.println("The two strings are the same.")
}
Option 2: String comparison with the equalsIgnoreCase method
In some string comparison tests you'll want to ignore whether the strings are uppercase or lowercase. When you want to test your strings for equality in this case-insensitive manner, use the equalsIgnoreCase method of the String class, like this:
String string1 = "foo";
String string2 = "FOO";
// java string compare while ignoring case
if (string1.equalsIgnoreCase(string2))
{
// this line WILL print
System.out.println("Ignoring case, the two strings are the same.")
}
Option 3: Java String comparison with the compareTo method
There is also a third, less common way to compare Java strings, and that's with the String class compareTo method. If the two strings are exactly the same, the compareTo method will return a value of 0 (zero). Here's a quick example of what this String comparison approach looks like:
String string1 = "foo bar";
String string2 = "foo bar";
// java string compare example
if (string1.compareTo(string2) == 0)
{
// this line WILL print
System.out.println("The two strings are the same.")
}
While I'm writing about this concept of equality in Java, it's important to note that the Java language includes an equals method in the base Java Object class. Whenever you're creating your own objects and you want to provide a means to see if two instances of your object are "equal", you should override (and implement) this equals method in your class (in the same way the Java language provides this equality/comparison behavior in the String equals method).
You may want to have a look at this ==, .equals(), compareTo(), and compare()
Function:
public float simpleSimilarity(String u, String v) {
String[] a = u.split(" ");
String[] b = v.split(" ");
long correct = 0;
int minLen = Math.min(a.length, b.length);
for (int i = 0; i < minLen; i++) {
String aa = a[i];
String bb = b[i];
int minWordLength = Math.min(aa.length(), bb.length());
for (int j = 0; j < minWordLength; j++) {
if (aa.charAt(j) == bb.charAt(j)) {
correct++;
}
}
}
return (float) (((double) correct) / Math.max(u.length(), v.length()));
}
Test:
String a = "This is the first string.";
String b = "this is not 1st string!";
// for exact string comparison, use .equals
boolean exact = a.equals(b);
// For similarity check, there are libraries for this
// Here I'll try a simple example I wrote
float similarity = simple_similarity(a,b);
The == operator check if the two references point to the same object or not. .equals() check for the actual string content (value).
Note that the .equals() method belongs to class Object (super class of all classes). You need to override it as per you class requirement, but for String it is already implemented, and it checks whether two strings have the same value or not.
Case 1
String s1 = "Stack Overflow";
String s2 = "Stack Overflow";
s1 == s2; //true
s1.equals(s2); //true
Reason: String literals created without null are stored in the String pool in the permgen area of heap. So both s1 and s2 point to same object in the pool.
Case 2
String s1 = new String("Stack Overflow");
String s2 = new String("Stack Overflow");
s1 == s2; //false
s1.equals(s2); //true
Reason: If you create a String object using the new keyword a separate space is allocated to it on the heap.
== compares the reference value of objects whereas the equals() method present in the java.lang.String class compares the contents of the String object (to another object).
I think that when you define a String you define an object. So you need to use .equals(). When you use primitive data types you use == but with String (and any object) you must use .equals().
If the equals() method is present in the java.lang.Object class, and it is expected to check for the equivalence of the state of objects! That means, the contents of the objects. Whereas the == operator is expected to check the actual object instances are same or not.
Example
Consider two different reference variables, str1 and str2:
str1 = new String("abc");
str2 = new String("abc");
If you use the equals()
System.out.println((str1.equals(str2))?"TRUE":"FALSE");
You will get the output as TRUE if you use ==.
System.out.println((str1==str2) ? "TRUE" : "FALSE");
Now you will get the FALSE as output, because both str1 and str2 are pointing to two different objects even though both of them share the same string content. It is because of new String() a new object is created every time.
Operator == is always meant for object reference comparison, whereas the String class .equals() method is overridden for content comparison:
String s1 = new String("abc");
String s2 = new String("abc");
System.out.println(s1 == s2); // It prints false (reference comparison)
System.out.println(s1.equals(s2)); // It prints true (content comparison)
All objects are guaranteed to have a .equals() method since Object contains a method, .equals(), that returns a boolean. It is the subclass' job to override this method if a further defining definition is required. Without it (i.e. using ==) only memory addresses are checked between two objects for equality. String overrides this .equals() method and instead of using the memory address it returns the comparison of strings at the character level for equality.
A key note is that strings are stored in one lump pool so once a string is created it is forever stored in a program at the same address. Strings do not change, they are immutable. This is why it is a bad idea to use regular string concatenation if you have a serious of amount of string processing to do. Instead you would use the StringBuilder classes provided. Remember the pointers to this string can change and if you were interested to see if two pointers were the same == would be a fine way to go. Strings themselves do not.
You can also use the compareTo() method to compare two Strings. If the compareTo result is 0, then the two strings are equal, otherwise the strings being compared are not equal.
The == compares the references and does not compare the actual strings. If you did create every string using new String(somestring).intern() then you can use the == operator to compare two strings, otherwise equals() or compareTo methods can only be used.
In Java, when the == operator is used to compare 2 objects, it checks to see if the objects refer to the same place in memory. In other words, it checks to see if the 2 object names are basically references to the same memory location.
The Java String class actually overrides the default equals() implementation in the Object class – and it overrides the method so that it checks only the values of the strings, not their locations in memory.
This means that if you call the equals() method to compare 2 String objects, then as long as the actual sequence of characters is equal, both objects are considered equal.
The == operator checks if the two strings are exactly the same object.
The .equals() method check if the two strings have the same value.
At the beginning of a method I want to check if the method is called with these exact parameters before, and if so, return the result that was returned back then.
At first, with one parameter, I used a Dictionary, but now I need to check 3 parameters (a String, an Object and a boolean).
I tried making a custom Object like so:
var cacheKey:Object = { identifier:identifier, type:type, someBoolean:someBoolean };
//if key already exists, return it (not working)
if (resultCache[cacheKey]) return resultCache[cacheKey];
//else: create result ...
//and save it in the cache
resultCache[cacheKey] = result;
But this doesn't work, because the seccond time the function is called, the new cacheKey is not the same object as the first, even though it's properties are the same.
So my question is: is there a datatype that will check the properties of the object used as key for a matching key?
And what else is my best option? Create a cache for the keys as well? :/
Note there are two aspects to the technical solution: equality comparison and indexing.
The Cliff Notes version:
It's easy to do custom equality comparison
In order to perform indexing, you need to know more than whether one object is equal to another -- you need to know which is object is "bigger" than the other.
If all of your properties are primitives you should squash them into a single string and use an Object to keep track of them (NOT a Dictionary).
If you need to compare some of the individual properties for reference equality you're going to have a write a function to determine which set of properties is bigger than the other, and then make your own collection class that uses the output of the comparison function to implement its own a binary search tree based indexing.
If the number of unique sets of arguments is in the several hundreds or less AND you do need reference comparison for your Object argument, just use an Array and the some method to do a naive comparison to all cached keys. Only you know how expensive your actual method is, so it's up to you to decide what lookup cost (which depends on the number of unique arguments provided to the function) is acceptable.
Equality comparison
To address equality comparison it is easy enough to write some code to compare objects for the values of their properties, rather than for reference equality. The following function enforces strict set comparison, so that both objects must contain exactly the same properties (no additional properties on either object allowed) with the same values:
public static propsEqual(obj1:Object, obj2:Object):Boolean {
for(key1:* in obj1) {
if(obj2[key1] === undefined)
return false;
if(obj2[key1] != obj2[key1])
return false;
}
for(key2:* in obj2)
if(obj1[key2] === undefined)
return false;
return true;
}
You could speed it up by eliminating the second for loop with the tradeoff that {A:1, B:2} will be deemed equal to {A:1, B:2, C:'An extra property'}.
Indexing
The problem with this in your case is that you lose the indexing that a Dictionary provides for reference equality or that an Object provides for string keys. You would have to compare each new set of function arguments to the entire list of previously seen arguments, such as using Array.some. I use the field currentArgs and the method to avoid generating a new closure every time.
private var cachedArgs:Array = [];
private var currentArgs:Object;
function yourMethod(stringArg:String, objArg:Object, boolArg:Boolean):* {
currentArgs = { stringArg:stringArg, objArg:objArg, boolArg:boolArg };
var iveSeenThisBefore:Boolean = cachedArgs.some(compareToCurrent);
if(!iveSeenThisBefore)
cachedArgs.push(currentArgs);
}
function compareToCurrent(obj:Object):Boolean {
return someUtil.propsEqual(obj, currentArgs);
}
This means comparison will be O(n) time, where n is the ever increasing number of unique sets of function arguments.
If all the arguments to your function are primitive, see the very similar question In AS3, where do you draw the line between Dictionary and ArrayCollection?. The title doesn't sound very similar but the solution in the accepted answer (yes I wrote it) addresses the exact same techinical issue -- using multiple primitive values as a single compound key. The basic gist in your case would be:
private var cachedArgs:Object = {};
function yourMethod(stringArg:String, objArg:Object, boolArg:Boolean):* {
var argKey:String = stringArg + objArg.toString() + (boolArg ? 'T' : 'F');
if(cachedArgs[argKey] === undefined)
cachedArgs[argKey] = _yourMethod(stringArg, objArg, boolArg);
return cachedArgs[argKey];
}
private function _yourMethod(stringArg:String, objArg:Object, boolArg:Boolean):* {
// Do stuff
return something;
}
If you really need to determine which reference is "bigger" than another (as the Dictionary does internally) you're going to have to wade into some ugly stuff, since Adobe has not yet provided any API to retrieve the "value" / "address" of a reference. The best thing I've found so far is this interesting hack: How can I get an instance's "memory location" in ActionScript?. Without doing a bunch of performance tests I don't know if using this hack to compare references will kill the advantages gained by binary search tree indexnig. Naturally it would depend on the number of keys.
I will run a set of experiments. The main method evaluated has the following signature:
[Model threshold] = detect(...
TrainNeg, TrainPos, nf, nT, factors, ...
removeEachStage, applyEstEachStage, removeFeatures);
where removeEachStage, applyEstEachStage, and removeFeatures are booleans. You can see that if I reverse the order of any of these boolean parameters I may get wrong results.
Is there a method in MATLAB that allows better organization in order to minimize this kind of error? Or is there any tool I can use to protect me against these errors?
Organization with a struct
You could input a struct that has these parameters as it's fields.
For example a structure with fields
setts.TrainNeg
.TrainPos
.nf
.nT
.factors
.removeEachStage
.applyEstEachStage
.removeFeatures
That way when you set the fields it is clear what the field is, unlike a function call where you have to remember the order of the parameters.
Then your function call becomes
[Model threshold] = detect(setts);
and your function definition would be something like
function [model, threshold] = detect(setts)
Then simply replace the occurrences of e.g. param with setts.param.
Mixed approach
You can also mix this approach with your current one if you prefer, e.g.
[Model threshold] = detect(in1, in2, setts);
if you wanted to still explicitly include in1 and in2, and bundle the rest into setts.
OOP approach
Another option is to turn detect into a class. The benefit to this is that a detect object would then have member variables with fixed names, as opposed to structs where if you make a typo when setting a field you just create a new field with the misspelled name.
For example
classdef detect()
properties
TrainNeg = [];
TrainPos = [];
nf = [];
nT = [];
factors = [];
removeEachStage = [];
applyEstEachStage = [];
removeFeatures =[];
end
methods
function run(self)
% Put the old detect code in here, use e.g. self.TrainNeg to access member variables (aka properties)
end
end
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Closed 10 years ago.
When learning a new programming language, one of the possible roadblocks you might encounter is the question whether the language is, by default, pass-by-value or pass-by-reference.
So here is my question to all of you, in your favorite language, how is it actually done? And what are the possible pitfalls?
Your favorite language can, of course, be anything you have ever played with: popular, obscure, esoteric, new, old...
Here is my own contribution for the Java programming language.
first some code:
public void swap(int x, int y)
{
int tmp = x;
x = y;
y = tmp;
}
calling this method will result in this:
int pi = 3;
int everything = 42;
swap(pi, everything);
System.out.println("pi: " + pi);
System.out.println("everything: " + everything);
"Output:
pi: 3
everything: 42"
even using 'real' objects will show a similar result:
public class MyObj {
private String msg;
private int number;
//getters and setters
public String getMsg() {
return this.msg;
}
public void setMsg(String msg) {
this.msg = msg;
}
public int getNumber() {
return this.number;
}
public void setNumber(int number) {
this.number = number;
}
//constructor
public MyObj(String msg, int number) {
setMsg(msg);
setNumber(number);
}
}
public static void swap(MyObj x, MyObj y)
{
MyObj tmp = x;
x = y;
y = tmp;
}
public static void main(String args[]) {
MyObj x = new MyObj("Hello world", 1);
MyObj y = new MyObj("Goodbye Cruel World", -1);
swap(x, y);
System.out.println(x.getMsg() + " -- "+ x.getNumber());
System.out.println(y.getMsg() + " -- "+ y.getNumber());
}
"Output:
Hello world -- 1
Goodbye Cruel World -- -1"
thus it is clear that Java passes its parameters by value, as the value for pi and everything and the MyObj objects aren't swapped.
be aware that "by value" is the only way in java to pass parameters to a method. (for example a language like c++ allows the developer to pass a parameter by reference using '&' after the parameter's type)
now the tricky part, or at least the part that will confuse most of the new java developers: (borrowed from javaworld)
Original author: Tony Sintes
public void tricky(Point arg1, Point arg2)
{
arg1.x = 100;
arg1.y = 100;
Point temp = arg1;
arg1 = arg2;
arg2 = temp;
}
public static void main(String [] args)
{
Point pnt1 = new Point(0,0);
Point pnt2 = new Point(0,0);
System.out.println("X: " + pnt1.x + " Y: " +pnt1.y);
System.out.println("X: " + pnt2.x + " Y: " +pnt2.y);
System.out.println(" ");
tricky(pnt1,pnt2);
System.out.println("X: " + pnt1.x + " Y:" + pnt1.y);
System.out.println("X: " + pnt2.x + " Y: " +pnt2.y);
}
"Output
X: 0 Y: 0
X: 0 Y: 0
X: 100 Y: 100
X: 0 Y: 0"
tricky successfully changes the value of pnt1!
This would imply that Objects are passed by reference, this is not the case!
A correct statement would be: the Object references are passed by value.
more from Tony Sintes:
The method successfully alters the
value of pnt1, even though it is
passed by value; however, a swap of
pnt1 and pnt2 fails! This is the major
source of confusion. In the main()
method, pnt1 and pnt2 are nothing more
than object references. When you pass
pnt1 and pnt2 to the tricky() method,
Java passes the references by value
just like any other parameter. This
means the references passed to the
method are actually copies of the
original references. Figure 1 below
shows two references pointing to the
same object after Java passes an
object to a method.
(source: javaworld.com)
Conclusion or a long story short:
Java passes it parameters by value
"by value" is the only way in java to pass a parameter to a method
using methods from the object given as parameter will alter the object as the references point to the original objects. (if that method itself alters some values)
useful links:
http://www.javaworld.com/javaworld/javaqa/2000-05/03-qa-0526-pass.html
http://www.ibm.com/developerworks/java/library/j-passbyval/
http://www.ibm.com/developerworks/library/j-praxis/pr1.html
http://javadude.com/articles/passbyvalue.htm
Here is another article for the c# programming language
c# passes its arguments by value (by default)
private void swap(string a, string b) {
string tmp = a;
a = b;
b = tmp;
}
calling this version of swap will thus have no result:
string x = "foo";
string y = "bar";
swap(x, y);
"output:
x: foo
y: bar"
however, unlike java c# does give the developer the opportunity to pass parameters by reference, this is done by using the 'ref' keyword before the type of the parameter:
private void swap(ref string a, ref string b) {
string tmp = a;
a = b;
b = tmp;
}
this swap will change the value of the referenced parameter:
string x = "foo";
string y = "bar";
swap(x, y);
"output:
x: bar
y: foo"
c# also has a out keyword, and the difference between ref and out is a subtle one.
from msdn:
The caller of a method which takes an
out parameter is not required to
assign to the variable passed as the
out parameter prior to the call;
however, the callee is required to
assign to the out parameter before
returning.
and
In contrast ref parameters are
considered initially assigned by the
callee. As such, the callee is not
required to assign to the ref
parameter before use. Ref parameters
are passed both into and out of a
method.
a small pitfall is, like in java, that objects passed by value can still be changed using their inner methods
conclusion:
c# passes its parameters, by default, by value
but when needed parameters can also be passed by reference using the ref keyword
inner methods from a parameter passed by value will alter the object (if that method itself alters some values)
useful links:
http://msdn.microsoft.com/en-us/vcsharp/aa336814.aspx
http://www.c-sharpcorner.com/UploadFile/saragana/Willswapwork11162005012542AM/Willswapwork.aspx
http://en.csharp-online.net/Value_vs_Reference
Python uses pass-by-value, but since all such values are object references, the net effect is something akin to pass-by-reference. However, Python programmers think more about whether an object type is mutable or immutable. Mutable objects can be changed in-place (e.g., dictionaries, lists, user-defined objects), whereas immutable objects can't (e.g., integers, strings, tuples).
The following example shows a function that is passed two arguments, an immutable string, and a mutable list.
>>> def do_something(a, b):
... a = "Red"
... b.append("Blue")
...
>>> a = "Yellow"
>>> b = ["Black", "Burgundy"]
>>> do_something(a, b)
>>> print a, b
Yellow ['Black', 'Burgundy', 'Blue']
The line a = "Red" merely creates a local name, a, for the string value "Red" and has no effect on the passed-in argument (which is now hidden, as a must refer to the local name from then on). Assignment is not an in-place operation, regardless of whether the argument is mutable or immutable.
The b parameter is a reference to a mutable list object, and the .append() method performs an in-place extension of the list, tacking on the new "Blue" string value.
(Because string objects are immutable, they don't have any methods that support in-place modifications.)
Once the function returns, the re-assignment of a has had no effect, while the extension of b clearly shows pass-by-reference style call semantics.
As mentioned before, even if the argument for a is a mutable type, the re-assignment within the function is not an in-place operation, and so there would be no change to the passed argument's value:
>>> a = ["Purple", "Violet"]
>>> do_something(a, b)
>>> print a, b
['Purple', 'Violet'] ['Black', 'Burgundy', 'Blue', 'Blue']
If you didn't want your list modified by the called function, you would instead use the immutable tuple type (identified by the parentheses in the literal form, rather than square brackets), which does not support the in-place .append() method:
>>> a = "Yellow"
>>> b = ("Black", "Burgundy")
>>> do_something(a, b)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 3, in do_something
AttributeError: 'tuple' object has no attribute 'append'
Since I haven't seen a Perl answer yet, I thought I'd write one.
Under the hood, Perl works effectively as pass-by-reference. Variables as function call arguments are passed referentially, constants are passed as read-only values, and results of expressions are passed as temporaries. The usual idioms to construct argument lists by list assignment from #_, or by shift tend to hide this from the user, giving the appearance of pass-by-value:
sub incr {
my ( $x ) = #_;
$x++;
}
my $value = 1;
incr($value);
say "Value is now $value";
This will print Value is now 1 because the $x++ has incremented the lexical variable declared within the incr() function, rather than the variable passed in. This pass-by-value style is usually what is wanted most of the time, as functions that modify their arguments are rare in Perl, and the style should be avoided.
However, if for some reason this behaviour is specifically desired, it can be achieved by operating directly on elements of the #_ array, because they will be aliases for variables passed into the function.
sub incr {
$_[0]++;
}
my $value = 1;
incr($value);
say "Value is now $value";
This time it will print Value is now 2, because the $_[0]++ expression incremented the actual $value variable. The way this works is that under the hood #_ is not a real array like most other arrays (such as would be obtained by my #array), but instead its elements are built directly out of the arguments passed to a function call. This allows you to construct pass-by-reference semantics if that would be required. Function call arguments that are plain variables are inserted as-is into this array, and constants or results of more complex expressions are inserted as read-only temporaries.
It is however exceedingly rare to do this in practice, because Perl supports reference values; that is, values that refer to other variables. Normally it is far clearer to construct a function that has an obvious side-effect on a variable, by passing in a reference to that variable. This is a clear indication to the reader at the callsite, that pass-by-reference semantics are in effect.
sub incr_ref {
my ( $ref ) = #_;
$$ref++;
}
my $value = 1;
incr(\$value);
say "Value is now $value";
Here the \ operator yields a reference in much the same way as the & address-of operator in C.
There's a good explanation here for .NET.
A lot of people are surprise that reference objects are actually passed by value (in both C# and Java). It's a copy of a stack address. This prevents a method from changing where the object actually points to, but still allows a method to change the values of the object. In C# its possible to pass a reference by reference, which means you can change where an actual object points to.
Don't forget there is also pass by name, and pass by value-result.
Pass by value-result is similar to pass by value, with the added aspect that the value is set in the original variable that was passed as the parameter. It can, to some extent, avoid interference with global variables. It is apparently better in partitioned memory, where a pass by reference could cause a page fault (Reference).
Pass by name means that the values are only calculated when they are actually used, rather than at the start of the procedure. Algol used pass-by-name, but an interesting side effect is that is it very difficult to write a swap procedure (Reference). Also, the expression passed by name is re-evaluated each time it is accessed, which can also have side effects.
Whatever you say as pass-by-value or pass-by-reference must be consistent across languages. The most common and consistent definition used across languages is that with pass-by-reference, you can pass a variable to a function "normally" (i.e. without explicitly taking address or anything like that), and the function can assign to (not mutate the contents of) the parameter inside the function and it will have the same effect as assigning to the variable in the calling scope.
From this view, the languages are grouped as follows; each group having the same passing semantics. If you think that two languages should not be put in the same group, I challenge you to come up with an example that distinguishes them.
The vast majority of languages including C, Java, Python, Ruby, JavaScript, Scheme, OCaml, Standard ML, Go, Objective-C, Smalltalk, etc. are all pass-by-value only. Passing a pointer value (some languages call it a "reference") does not count as pass by reference; we are only concerned about the thing passed, the pointer, not the thing pointed to.
Languages such as C++, C#, PHP are by default pass-by-value like the languages above, but functions can explicitly declare parameters to be pass-by-reference, using & or ref.
Perl is always pass-by-reference; however, in practice people almost always copy the values after getting it, thus using it in a pass-by-value way.
by value
is slower than by reference since the system has to copy the parameter
used for input only
by reference
faster since only a pointer is passed
used for input and output
can be very dangerous if used in conjunction with global variables
Concerning J, while there is only, AFAIK, passing by value, there is a form of passing by reference which enables moving a lot of data. You simply pass something known as a locale to a verb (or function). It can be an instance of a class or just a generic container.
spaceused=: [: 7!:5 <
exectime =: 6!:2
big_chunk_of_data =. i. 1000 1000 100
passbyvalue =: 3 : 0
$ y
''
)
locale =. cocreate''
big_chunk_of_data__locale =. big_chunk_of_data
passbyreference =: 3 : 0
l =. y
$ big_chunk_of_data__l
''
)
exectime 'passbyvalue big_chunk_of_data'
0.00205586720663967
exectime 'passbyreference locale'
8.57957102144893e_6
The obvious disadvantage is that you need to know the name of your variable in some way in the called function. But this technique can move a lot of data painlessly. That's why, while technically not pass by reference, I call it "pretty much that".
PHP is also pass by value.
<?php
class Holder {
private $value;
public function __construct($value) {
$this->value = $value;
}
public function getValue() {
return $this->value;
}
}
function swap($x, $y) {
$tmp = $x;
$x = $y;
$y = $tmp;
}
$a = new Holder('a');
$b = new Holder('b');
swap($a, $b);
echo $a->getValue() . ", " . $b->getValue() . "\n";
Outputs:
a b
However in PHP4 objects were treated like primitives. Which means:
<?php
$myData = new Holder('this should be replaced');
function replaceWithGreeting($holder) {
$myData->setValue('hello');
}
replaceWithGreeting($myData);
echo $myData->getValue(); // Prints out "this should be replaced"
By default, ANSI/ISO C uses either--it depends on how you declare your function and its parameters.
If you declare your function parameters as pointers then the function will be pass-by-reference, and if you declare your function parameters as not-pointer variables then the function will be pass-by-value.
void swap(int *x, int *y); //< Declared as pass-by-reference.
void swap(int x, int y); //< Declared as pass-by-value (and probably doesn't do anything useful.)
You can run into problems if you create a function that returns a pointer to a non-static variable that was created within that function. The returned value of the following code would be undefined--there is no way to know if the memory space allocated to the temporary variable created in the function was overwritten or not.
float *FtoC(float temp)
{
float c;
c = (temp-32)*9/5;
return &c;
}
You could, however, return a reference to a static variable or a pointer that was passed in the parameter list.
float *FtoC(float *temp)
{
*temp = (*temp-32)*9/5;
return temp;
}