How can I pass strings by reference to the parent scope?
This doesn't work since strings are not acceptable "values".
function Submit([ref]$firstName){
$firstName.value = $txtFirstName.Text
}
$firstName = $null
Submit([ref]$firstName)
$firstName
Error: "Property 'value' cannot be found on this object; make sure it exists and is settable"
Doing this doesn't give an error but it doesn't change the variable either:
$firstName = "nothing"
function Submit([ref]$firstName){
$firstName = $txtFirstName.Text
}
Submit([ref]$firstName)
$firstName
Edit:
Doing the first code block by itself works. However when trying to do it in my script it returns the error again. I fixed it enough for it to assign the variable and do what I want but it still throws up an error and I was wondering how to fix that. I think it's because it doesn't like variable;es changing during a running session. Here is a link to my script
https://github.com/InconspicuousIntern/Form/blob/master/Form.ps1
Your first snippet is conceptually correct and works as intended - by itself it does not produce the "Property 'Value' cannot be found on this object" error.
You're seeing the error only as part of the full script you link to, because of the following line:
$btnSubmit.Add_Click({ Submit })
This line causes your Submit function to be called without arguments, which in turn causes the $firstName parameter value to be $null, which in turn causes the error quoted above when you assign to $firstName.Value.
By contrast, the following invocation of Submit, as in your first snippet, is correct:
Submit ([ref] $firstName) # Note the recommended space after 'Submit' - see below.
[ref] $firstName creates a (transient) reference to the caller's $firstName variable, which inside Submit binds to (local) parameter variable $firstName (the two may, but needn't and perhaps better not have the same name), where $firstName.Value can then be used to modify the caller's $firstName variable.
Syntax note: I've intentionally placed a space between Submit and ([ref] $firstName) to make one thing clearer:
The (...) (parentheses) here do not enclose the entire argument list, as they would in a method call, they enclose the single argument [ref] $firstName - of necessity, because that expression wouldn't be recognized as such otherwise.
Function calls in PowerShell are parsed in so-called argument mode, whose syntax is more like that of invoking console applications: arguments are space-separated, and generally only need quoting if they contain special characters.
For instance, if you also wanted to pass string 'foo', as the 2nd positional parameter, to Submit:
Submit ([ref] $firstName) foo
Note how the two arguments are space-separated and how foo needn't be quoted.
As for an alternative approach:
[ref]'s primary purpose is to enable .NET method calls that have ref / out parameters, and, as shown above, using [ref] is nontrivial.
For calls to PowerShell functions there are generally simpler solutions.
For instance, you can pass a custom object to your function and let the function update its properties with the values you want to return, which naturally allows multiple values to be "returned"; e.g.:
function Submit($outObj){
$outObj.firstName = 'a first name'
}
# Initialize the custom object that will receive values inside
# the Submit function.
$obj = [pscustomobject] #{ firstName = $null }
# Pass the custom object to Submit.
# Since a custom object is a reference type, a *reference* to it
# is bound to the $outObj parameter variable.
Submit $obj
$obj.firstName # -> 'a first name'
Alternatively, you can just let Submit construct the custom object itself, and simply output it:
function Submit {
# Construct and (implicitly) output a custom
# object with all values of interest.
[pscustomobject] #{
firstName = 'a first name'
}
}
$obj = Submit
$obj.firstName # -> 'a first name'
Please try this out and see if you are getting the same results? It is working for me, and I really did not change much.
$txtFirstName = [PSCustomObject]#{
Text = "Something"
}
function Submit([ref]$var){
$var.value = $txtFirstName.Text
}
$firstName = $null
Submit([ref]$firstName)
$firstName
I'm trying to write a custom validateParameter function with Perl.
I have the following code which also works:
sub validateParameter {
my ($args, $list) = #_;
if ( ref($list) eq "ARRAY" ) {
foreach my $key (#$list) {
if ( not defined $args->{$key} ) {
die "no $key given!";
}
}
}
#elsif ( check if string ) {
#}
}
I want to call my function the following way:
validateParameter({ hallo => "Welt", test => "Blup"}, ["hallo", "test"]);
But I also want to call my function like this:
validateParameter({ hallo => "Welt", test => "Blup"}, "hallo");
I know that Perl only has the following three data-types (scalars, hashes, arrays). But maybe there is a smart way to check if a variable is a string.
How can I check if the given arg is a string?
Update: I somehow missed the end of the question. Just testing ref($list) eq 'ARRAY' will work most of the time, but to properly allow even overloaded objects, you should just try dereferencing the parameter:
if ( eval { \#$list } ) {
# it was an array
}
else {
# assume it is a string
}
Original answer:
You can check a number of things about a parameter:
if ( ! defined $param ) {
# undefined
}
elsif ( defined Scalar::Util::blessed($param) ) {
# object
}
elsif ( ref $param ) {
# reference (return of ref will give the type)
}
elsif ( length do { no warnings "numeric"; $param & '' } ) {
# number
}
else {
# string
}
But all of that (except perhaps the defined check) kind of defeats the purpose of Perl's automatically converting to your desired type and will limit what can be passed (for instance, a string or dualvar where a number is wanted, or an overloaded object where a string, number, or reference is wanted, or tied variables or magic variables such as $!).
You may want to also just look at what Params::Validate can do.
Don't base behaviour on the "type" of arguments because there really isn't such a thing. You will run into problems if you use type-base polymorphism because Perl values often have more than one type.
For example,
The scalar produced by "123" is stored as as string, but Perl doesn't distinguish it from the scalar produced by 123 which isn't stored as a string.
Scalars can contain both a number and a cached stringification of that number. (e.g. my $i = 123; "".$i;)
Scalars can contain both a number and a string (that isn't a stringification of the number). Common examples of these are $! and !1.
A reference to an object that overloads stringification is also a useful string. (e.g. DateTime->now)
A reference to an array may overload %{} to usable as a hash reference. A reference to an hash may overload #{} to usable as an array reference.
And more.
No, there is no way to check if a scalar is a string, as Perl does implicit type conversions depending on context. If you give a number as the second argument to your function, and you use it in a context that requires a string, it will be automatically converted to a string. So, just check if ref($list) is empty - in such case, $list is not a reference, and therefore it is a string or a number, and you don't need to distinguish between those two.
Im trying to modify a variable using upvar (in an upward stack), but the value of the variable is passed to the procedure and not the variable name.
I cannot change what is passed, since it is already implemented widely on the program.
Is there a way to modify the file name in some way ?
proc check_file_exists {name} {
upvar $name newName
check_exists $name #Do all checks that file is there
set newName $name_1
}
check_file_exists $name
puts $name
this code will print the old name of the file and not the new one.
What I think you should do is bite the bullet and change the calls. It's a fairly simple search-and-replace, after all. The code will be more sane than if you use any of the other solutions.
check_file_exists name
Or, you could add another parameter to the parameter list and use that to pass the name, making the first argument a dummy argument.
check_file_exists $name name
Or, if you're not using the return value, you could return the new value and assign it back:
set name [check_file_exists $name]
Or, you could assign the new value to a global variable (e.g. theValue) inside the procedure, and assign that back:
check_file_exists $name
# don't need this if you're in global scope
global theValue
set name $theValue
Or, you could assign the name to a global variable (e.g. theName) and access that inside the procedure: the procedure will be able to update name directly.
# don't need this if you're in global scope
global theName
set theName name
check_file_exists $name
(There are some variations on this f.i. using upvar.)
None of the alternatives are pretty, and all of them still require you to make a change at the call (except the last, if you only ever use one variable for this value). If you're adamant about not doing that, there's always Donal's info frame solution, which only requires the procedure itself to be changed.
Let me know if you want help with the procedure code for any of these alternatives.
This is quite difficult; it's really not the way you're supposed to work. But you can do it using info frame -1 (a facility usually used for debugging) to find out exactly how the current procedure was called. However, you need to be careful as the caller might be using the result of a command: this is an unsafe hack.
proc check_file_exists {name} {
set caller [dict get [info frame -1] cmd]
if {[regexp {^check_file_exists +\$(\w+)} $caller -> varName]} {
# OK, we were called with a simple variable name
puts "Called with variable $varName"
} else {
# Complicated case! Help...
return -code error "must be called with a simple variable's contents"
}
upvar 1 $varName newName
check_exists $name
set newName $name_1
}
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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;
}