Pass arguments from a function to another function

Question

I have something like this

  function f1($name = ''){
    f2("prefix_{$name}", $args);
  }

the f1() function is called like:

f1('name', $var1, $var2, $var3);

How can I pass these variables to f2() inside it, the same way they are passed to f1() ?

so instead of f2("prefix_{$name}", $args);

it should be like f2("prefix_{$name}", $var1, $var2, $var3);

Not that I have no control over the f2 function :(

Community · Accepted Answer · 2017-05-23 10:34:22Z

2

function f1( $name = '' )
{
    // get all arguments to this function
    $args = func_get_args();
    // prefix the first arguments with some prefix
    $args[ 0 ] = 'prefix_' . $name;
    // call the second function with the $args array as its arguments
    call_user_func_array( 'f2', $args );
}

edit:
replaced

$args[ 0 ] = 'prefix_' . $args[ 0 ];

with

$args[ 0 ] = 'prefix_' . $name;

to account for what Gerry said in his answer.

edited May 23, 2017 at 10:34

CommunityBot

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answered Aug 14, 2011 at 1:04

Decent Dabbler

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Comments

Gerry · Accepted Answer · 2011-08-14 00:54:18Z

1

The regular func_get_args() will not take into account any default arguments, so you will need to do something like this:

function func_get_default_args($a) {
    $args = array_slice(func_get_args(), 1);
    return array_merge($args, array_slice($a, sizeof($args)));
}

function f1($name = ''){
    call_user_func_array('f2', func_get_default_args(func_get_args(), $name));
}

answered Aug 14, 2011 at 0:54

Gerry

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