Passing an array as a reference to a function

[Bit Byte]

Noobie here (C++/C/Java experience though) ...

Recently picked up PHP ...

I want to pass an array to a function and then to use count on the
passed variable - is this the way to do it (its an object method) :

public function foo(array& $arr)
{
if (count($arr) < 1)
throw new MyException("Bl ah blah ...") ;
....
}


I have two questions for the pros:

1). Is the function signature correct (valid syntax?)
2). Can count accept a reference type variable ?

[countach]

Re: Passing an array as a reference to a function

En las nuevas, el Bit Byte escribió:

Noobie here (C++/C/Java experience though) ...
>
Recently picked up PHP ...
>
I want to pass an array to a function and then to use count on the
passed variable - is this the way to do it (its an object method) :
>
public function foo(array& $arr)
{
if (count($arr) < 1)
throw new MyException("Bl ah blah ...") ;
....
}
>
>
I have two questions for the pros:
>
1). Is the function signature correct (valid syntax?)
2). Can count accept a reference type variable ?

1) Correct syntax:

public function foo(&$arr)
{
if (count($arr) < 1)
throw new MyException("Bl ah blah ...") ;
....
}

2) Yes

[Robin]

Re: Passing an array as a reference to a function

Bit Byte wrote:

1). Is the function signature correct (valid syntax?)

Depends on your PHP-Version, i.e.:

PHP 4:
function foo(&$arr){}

PHP 4 knows nothing about visibilty and always passes by value, hence
the '&' to signal a pass-by-reference. Nice text about using objects
and references in PHP 4:

http://www.obdev.at/developers/articles/00002.html

PHP 5:
public function foo(array $arr){}

PHP 5 always passes by reference, so no '&' is needed.

Cheers
Robin

[Colin Fine]

Re: Passing an array as a reference to a function

Robin wrote:

Bit Byte wrote:

>1). Is the function signature correct (valid syntax?)

>
Depends on your PHP-Version, i.e.:
>
PHP 4:
function foo(&$arr){}
>
PHP 4 knows nothing about visibilty and always passes by value, hence
the '&' to signal a pass-by-reference. Nice text about using objects
and references in PHP 4:

http://www.obdev.at/developers/articles/00002.html

>
PHP 5:
public function foo(array $arr){}
>
PHP 5 always passes by reference, so no '&' is needed.
>
Cheers
Robin

I don't believe this is correct.

http://www.php.net/manual/en/functions.arguments.php says:

"By default, function arguments are passed by value (so that if you
change the value of the argument within the function, it does not get
changed outside of the function). If you wish to allow a function to
modify its arguments, you must pass them by reference.

"If you want an argument to a function to always be passed by reference,
you can prepend an ampersand (&) to the argument name in the function
definition:"

I think that what you are thinking of is that in PHP5 *objects* are
always passed by reference, but arrays are not objects.

Therefore the & *is* still needed in PHP5.

Furthermore, I can't find where it says so, but I'm pretty sure that the
PHP5 extension that allows you to specify a type for a function argument
is limited to objects.

If I am right, then :

for an array in PHP5

public function foo(&$arr) {

but if you use an object class MyArray

public function foo (MyArray $arr) {

Colin

[Bit Byte]

Re: Passing an array as a reference to a function



Colin Fine wrote:

Robin wrote:
>

>Bit Byte wrote:
>>

>>1). Is the function signature correct (valid syntax?)

>>
>>
>Depends on your PHP-Version, i.e.:
>>
>PHP 4:
>function foo(&$arr){}
>>
>PHP 4 knows nothing about visibilty and always passes by value, hence
>the '&' to signal a pass-by-reference. Nice text about using objects
>and references in PHP 4:
>http://www.obdev.at/developers/articles/00002.html
>>
>PHP 5:
>public function foo(array $arr){}
>>
>PHP 5 always passes by reference, so no '&' is needed.
>>
>Cheers
>Robin

>
>
I don't believe this is correct.
>
http://www.php.net/manual/en/functions.arguments.php says:
>
"By default, function arguments are passed by value (so that if you
change the value of the argument within the function, it does not get
changed outside of the function). If you wish to allow a function to
modify its arguments, you must pass them by reference.
>
"If you want an argument to a function to always be passed by reference,
you can prepend an ampersand (&) to the argument name in the function
definition:"
>
I think that what you are thinking of is that in PHP5 *objects* are
always passed by reference, but arrays are not objects.
>
Therefore the & *is* still needed in PHP5.
>
Furthermore, I can't find where it says so, but I'm pretty sure that the
PHP5 extension that allows you to specify a type for a function argument
is limited to objects.
>
If I am right, then :
>
for an array in PHP5
>
public function foo(&$arr) {
>
but if you use an object class MyArray
>
public function foo (MyArray $arr) {
>
Colin
>
>
>

Thanks all. This is a very detailed (and very useful) reply.