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up vote 4 down vote favorite
1

I have really simple problem in my PHP script. There is a function defined which takes variable length argument list:

function foo() {
  // func_get_args() and similar stuff here
}

When I call it like this, it works just fine:

foo("hello", "world");

However, I have my variables in the array and I need to pass them "separately" as single arguments to the function. For example:

$my_args = array("hello", "world");
foo(do_some_stuff($my_args));

Is there any do_some_stuff function which splits the arguments for me so I can pass them to the function?

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7 Answers

up vote 9 down vote accepted

Use ReflectionFunction::invokeArgs(array $args) or call_user_func_array( callback $callback , array $param_arr).

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+1 for Reflections. Although it's like an overkill here, but it's always cool to deal with them! –  dfsq Jan 28 '12 at 13:00
 
Thanks, both your solutions solve my problem perfectly. –  Pavel S. Jan 28 '12 at 13:08
up vote 4 down vote

Well you need call_user_func_array

call_user_func_array('foo', $my_args);

http://php.net/manual/en/function.call-user-func-array.php

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Thanks for the answer, it's exactly what I need. However, I can accept only one answer. –  Pavel S. Jan 28 '12 at 13:03
up vote 2 down vote

Sounds to me like you are looking for call_user_func_array.

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Thanks for the answer, it's exactly what I need. However, I can accept only one answer. –  Pavel S. Jan 28 '12 at 12:59
 
Just accept the one that either gives the most required detail, or the earliest one :) –  Cags Jan 28 '12 at 13:05
up vote 1 down vote

You are searching for call_user_func_array().

http://it2.php.net/manual/en/function.call-user-func-array.php

Usage:

$my_args = array("hello", "world");
call_user_func_array('foo', $my_args);

// Equivalent to:
foo("hello", "world");
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Thank you for the comparison –  jocken Oct 10 '12 at 13:41
up vote 0 down vote

http://www.php.net/manual/en/functions.arguments.php#functions.variable-arg-list

Isn't this what you want?

edit ah... ok... how about this: Passing an Array as Arguments, not an Array, in PHP

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Nope. I have already read this. All the functions are supposed to work inside my "foo" function. I need to split the arguments before they are passed to the function. –  Pavel S. Jan 28 '12 at 12:56
up vote 0 down vote

If you can change the code of foo() it should be easy to solve this in just one place.

function foo()
{
    $args = func_get_args();
    if(count($args) == 1 && is_array($args[0]))
    {
        $args = $args[0]
    }
    // use $args as normal
}
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This not works if he wants to pass an array (for real) as first argument. –  lorenzo-s Jan 28 '12 at 13:00
 
No, it does not work if there's only one argument and that argument is supposed to be an array. If there's more than one argument then they're all left alone. –  Arjan Jan 28 '12 at 13:04
 
Yup, I saw, I saw... –  lorenzo-s Jan 28 '12 at 13:05
up vote 0 down vote

This solution is not recommended at all, but just showing a possibility :

Using eval

eval ( "foo('" . implode("', '", $args_array) . "' )" );

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1  
I know it's only "showing a possibility", but still... it's eval. ;) –  Crozin Jan 28 '12 at 13:12

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