I am trying to understand the return type function but I really can't. For example in this code:
int add(int a, int b){
result=a-(-b);
return result;
}
I cant understand why it became return result and what is it for and what it do.I am new in c++ and Iwanted to become better in c++.
This is meant to give back an answer (value) to the other part of your program, which called this function.
Let's say you do something like:
print(add(2,4));
Your program would print the value: 6
Disclaimer: not really sure if the syntax is correct, it's been over a decade since I last programmed in C++;
Related
I've got program below:
#include<type_traits>
#include<iostream>
using namespace std;
template <class F, class R = typename result_of<F()>::type>
R call(F& f) { return f(); }
struct S {
double operator()(){return 0.0;}
};
int f(){return 1;}
int main()
{
S obj;
call(obj);//ok
call(f);//error!
return 0;
}
It fails to compile in the line of "call(f)".
It's weird that "call(obj)" is OK.
(1) I've a similar post in another thread C++11 result_of deducing my function type failed. But it doesn't tell why functor objects are OK while functions are not.
(2) I'm not sure if this is related to "R call(F& f)": a function type cannot declare a l-value?
(3) As long as I know, any token with a name, like variable/function, should be considered a l-value. And in the case of function parameter, compiler should "decay" my function name "f" to a function pointer, right?
(4) This is like decaying an array and pass it to a function----And a function pointer could be an l-value, then what's wrong with "call(F& f)"?
Would you help to give some further explanations on "why" is my case, where did I get wrong?
Thanks.
The problem with call(f) is that you deduce F as a function type, so it doesn't decay to a function pointer. Instead you get a reference to a function. Then the result_of<F()> expression is invalid, because F() is int()() i.e. a function that returns a function, which is not a valid type in C++ (functions can return pointers to functions, or references to functions, but not functions).
It will work if you use result_of<F&()> which is more accurate anyway, because that's how you're calling the callable object. Inside call(F& f) you do f() and in that context f is an lvalue, so you should ask what the result of invoking an lvalue F with no arguments is, otherwise you could get the wrong answer. Consider:
struct S {
double operator()()& {return 0.0;}
void operator()()&& { }
};
Now result_of<F()>::type is void, which is not the answer you want.
If you use result_of<F&()> then you get the right answer, and it also works when F is a function type, so call(f) works too.
(3) As long as I know, any token with a name, like variable/function, should be considered a l-value. And in the case of function parameter, compiler should "decay" my function name "f" to a function pointer, right?
No, see above. Your call(F&) function takes its argument by reference, so there is no decay.
(4) This is like decaying an array and pass it to a function----And a function pointer could be an l-value, then what's wrong with "call(F& f)"?
Arrays don't decay when you pass them by reference either.
If you want the argument to decay then you should write call(F f) not call(F& f). But even if you do that you still need to use result_of correctly to get the result of f() where f is an lvalue.
I am using the Arduino IDE to write code and am trying to understand the namespace stuff.
My thought is, is there a way to shorten the many places (in my code) where I have things like:
Serial.print("a="); Serial.print(a); Serial.print(" b="); Serial.println(b);
to something shorter like:
S.print(...
or
sprint(...
Can it be done?
I tried using String concatenation but it is very limited and expensive. That is just adding one
String s;
to my code at the global level increased the download size by 1482 bytes. And you can't do something like:
Serial.print("a=" + a); Serial.println(" b=" + b);
because it cant handle starting a concatenation with a literal string.
Any thoughts welcome.
Arduino uses the C++ language. It is not considered good practice, but you could use a preprocessor macro:
#define sprint Serial.print
You could use a pointer and member de-reference operator, like this:
HardwareSerial *my_device;
void setup()
{
my_device->begin(9600);
delay(100);
}
void loop()
{
if (my_device->available())
{
int r = my_device->read();
// etc.
}
}
void increment(ref int i)
{
++i;
}
class Class
{
immutable int member;
this(int parameter)
{
member = parameter;
++member; // okay
increment(member); // compile-time error
}
}
Why is ++member okay, but increment(member) isn't? Shouldn't both behave the same way?
Probably because the reference to increment isn't scope, so it has the potential to be escaped past the scope of the constructor, which would break the immutability of member, and the compiler can't verify that it's fine.
(It might be that scope won't work either, but it should. If implemented properly, I think scope would fix a lot of bugs like these, as well as providing for interesting optimizations. If it doesn't, I'd say it's a bug.)
I've pointed out semi-similar bugs before, but with delegates.
Const/immutable do have such problems in D.
What if increment was this?
int* p;
void increment(ref int i)
{
p = &i;
}
Uh oh, you've created a mutable reference to immutable data, breaking the type system.
I am guessing that
this(int parameter) {
member = parameter;
++member;
}
is an equivalent of
Class(int parameter): member(parameter+1) {}
in C++.
I think member field is not truly mutable in constructor, so compiler can optimize it to just init it. But it cannot do it with call to another function.
PS. It works on ideone: http://ideone.com/5ym5u
public function t()
{
if(xxx)return xxx;
//don't return anything
}
How to define the return type for such method?
A function either has to return nothing, or return something - it can't do both. The reason being, what if you write this code:
var someValue = someFunction();
How would this code be handled if sometimes someFunction returned a value, and sometimes it didn't?
Your problem is a really common one, though, and there are several ways to work around it.
Sentinel Values
You can return special-case values that you treat as non-values (such as null, NaN, "", or -1). These special-case values are called sentinel values. The users of your function (maybe your own code) would check the result after calling it. If it gets one of the sentinel values back, it doesn't use the result.
Try-style Functions
You could also use a pattern commonly used in C#/.Net that they call "Try-methods", but you can call "Try-functions". Actionscript doesn't work exactly like C#, but we can get close enough for this pattern to work.
To do this, return Boolean, returning true if you have a value, and false if you don't. Then give your function an Object parameter, and populate a property with the actual return value:
function trySomeFunction(result:Object) : Boolean
{
if(xxx)
{
result.Value = xxx;
return true;
}
return false;
}
// ...
var result:Object = { Value:null };
if(trySomeFunction(result))
{
// Do something with the value here
}
Exceptions
If your method failed to do what it promised to do, you can throw an exception. Then you don't have to worry about what gets returned, because your code just dies.
Exceptions are great because people who call your code don't have to learn as much. They don't have to know that your function only succeeded if it doesn't return some magic value that is different for every function, they don't need to check if your function returns true/false, and they don't need to know that some property gets populated on an object when the function succeeded.
Without exceptions, if they forget to check for your special value, or the result of a "Try-function", an error might happen further on in the program. Or it might not happen at all, and the program will continue running, but be broken. In these cases, it is much harder to track down the problem.
With exceptions, your code just blows up the second it detects a problem, and gives you the line of code where your program realized it couldn't work correctly.
If it is normal and okay for your function to fail, though, you probably shouldn't throw an exception. You should probably use a "Try-function" or a sentinel value instead.
Everything that Merlyn says is fine, though perhaps a bit of overkill. If your method needs the potential to return null, then just pass xxx back whether it's null or not...
public function t():MyReturnType
{
return xxx;
}
... since you're going to have to do check the special condition in the calling method anyway:
public function caller():void
{
var value:MyReturnType = t();
if (value)
doSomethingPositive();
else
copeWithNullState();
}
Some people think that this is wrong and advocate creating a special 'null value' object in your return class, like this:
public class MyReturnType
{
public static const NULL:MyReturnType = new MyReturnType(null);
public function MyReturnType(identifier:String)
...
}
then in your function either return an interesting MyReturnType or NULL:
public function t():MyReturnType
{
return xxx || MyReturnType.NULL;
}
but then you've not really improved your caller method so why bother?
public function caller():void
{
var value:MyReturnType = t();
if (value != MyReturnType.NULL)
doSomethingPositive();
else
copeWithNullState();
}
Whatever you choose to do, eventually you're just going to have to test for special cases in the caller method. Personally I'd say it's better to keep it as simple as possible, and the special 'null value' in your class is over-complication.
I write plenty of functions that might return an object reference, but may also return null, like this:
public function findThingById( id:int ):MyThingType
{
... code here to search for/load/lookup thing with matching id
if (found)
{
return thing;
}
return null;
}
Wherever you call a function that might return null, you would either test the return value for null explicitly, or if you expect null just don't do anything with the return value. For built-in data types, you would use a special value rather than null (like NaN for Numbers illustrated above), but the principle is the same.
public function t():type
{
if(xxx)return xxx;
//don't return anything
}
so something like:
private function derp():void{
}
private function derp2():int{
}
private function derp3():Boolean{
}
etc
its always good practice to define the return type, and return an indicator variable if false (i.e return -1 for number), but if its really necessary you can do this:
public function t()
{
if(xxx)return xxx;
//don't return anything
}
don't specify a return type at all, and either it will return xxx or undefined, but I highly encourage using a return type.
But since flash runs on virtual memory it can basically adapt to your return type so it makes it a little flexible so you can return what ever.
EDIT:
public function t()
{
if(true){
return xxx; //xxx can by of any type
}
return NaN
}
EDIT:
so lets say you want the condition with a return type of int you would have
public function t():int
{
if(true){
return xxx; //where xxx != -1
}
return -1
}
this when we get back any number other then negative 1 we know the condition was true, but if we get -1 the condition was false. But if you numbers are any real numbers, use NaN keyword:
public function t():Number
{
if(true){
return xxx; //where xxx != -1
}
return NaN
}
this way if you get NaN back it means your condition was false
I should also mention that when your dealing with NaN use the isNaN() to determine if the return type was NaN or not:
example:
if(isNaN(t()) == true){
//if condition in function t() was false;
}
There is no way a single function can return void as well as returning a type. This is because void is not part of any inheritance chain, it is a top level concept meaning something very specific; it doesn't return anything.
Don't get confused between null and void, undefined and void, NaN and void. They are different concepts.
The answer to your question is simply that it can't, nor should it be done. When you hit a problem like this, it's good practice to step back and ask 'why' do you need to do this.
I would highly encourage that you mark this question as closed, and instead posted a higher level problem. For example,
How would I structure a class that has a variable xxx that does this and that and will be eventually set to a number when the user clicks. But if the user never clicks, it will be unset.... ecetera ecetera....
If you are 100% convinced you should do something like you're asking, you could 'fake' it by
function f():* {
var xxx:Number = 999;
if(xxx) {
return xxx;
} else {
return undefined;
}
}
var xx:* = f();
trace(xx);
But remember, from that point onwards you will need to check everytime xx is used to see if it undefined. Otherwise it will lead to errors.
What does "Overloaded"/"Overload" mean in regards to programming?
It means that you are providing a function (method or operator) with the same name, but with a different signature.
For example:
void doSomething();
int doSomething(string x);
int doSomething(int a, int b, int c);
Basic Concept
Overloading, or "method overloading" is the name of the concept of having more than one methods with the same name but with different parameters.
For e.g. System.DateTime class in c# have more than one ToString method. The standard ToString uses the default culture of the system to convert the datetime to string:
new DateTime(2008, 11, 14).ToString(); // returns "14/11/2008" in America
while another overload of the same method allows the user to customize the format:
new DateTime(2008, 11, 14).ToString("dd MMM yyyy"); // returns "11 Nov 2008"
Sometimes parameter name may be the same but the parameter types may differ:
Convert.ToInt32(123m);
converts a decimal to int while
Convert.ToInt32("123");
converts a string to int.
Overload Resolution
For finding the best overload to call, compiler performs an operation named "overload resolution". For the first example, compiler can find the best method simply by matching the argument count. For the second example, compiler automatically calls the decimal version of replace method if you pass a decimal parameter and calls string version if you pass a string parameter. From the list of possible outputs, if compiler cannot find a suitable one to call, you will get a compiler error like "The best overload does not match the parameters...".
You can find lots of information on how different compilers perform overload resolution.
A function is overloaded when it has more than one signature. This means that you can call it with different argument types. For instance, you may have a function for printing a variable on screen, and you can define it for different argument types:
void print(int i);
void print(char i);
void print(UserDefinedType t);
In this case, the function print() would have three overloads.
It means having different versions of the same function which take different types of parameters. Such a function is "overloaded". For example, take the following function:
void Print(std::string str) {
std::cout << str << endl;
}
You can use this function to print a string to the screen. However, this function cannot be used when you want to print an integer, you can then make a second version of the function, like this:
void Print(int i) {
std::cout << i << endl;
}
Now the function is overloaded, and which version of the function will be called depends on the parameters you give it.
Others have answered what an overload is. When you are starting out it gets confused with override/overriding.
As opposed to overloading, overriding is defining a method with the same signature in the subclass (or child class), which overrides the parent classes implementation. Some language require explicit directive, such as virtual member function in C++ or override in Delphi and C#.
using System;
public class DrawingObject
{
public virtual void Draw()
{
Console.WriteLine("I'm just a generic drawing object.");
}
}
public class Line : DrawingObject
{
public override void Draw()
{
Console.WriteLine("I'm a Line.");
}
}
An overloaded method is one with several options for the number and type of parameters. For instance:
foo(foo)
foo(foo, bar)
both would do relatively the same thing but one has a second parameter for more options
Also you can have the same method take different types
int Convert(int i)
int Convert(double i)
int Convert(float i)
Just like in common usage, it refers to something (in this case, a method name), doing more than one job.
Overloading is the poor man's version of multimethods from CLOS and other languages. It's the confusing one.
Overriding is the usual OO one. It goes with inheritance, we call it redefinition too (e.g. in https://stackoverflow.com/users/3827/eed3si9n's answer Line provides a specialized definition of Draw().