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Home/ Questions/Q 3351100
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T01:52:03+00:00 2026-05-18T01:52:03+00:00

Consider the following functions: function debug() { $args = func_get_args(); // process $args }

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Consider the following functions:

function debug() {
  $args = func_get_args();
  // process $args
}

function debug_die() {
  // call debug() with the passed arguments
  die;
}

The method debug_die exits after calling debug that takes a variable number of arguments.

So the arguments passed to debug_die as such are meant for debug only and just have to be forwarded. How can this be done in the debug_die method?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-18T01:52:04+00:00Added an answer on May 18, 2026 at 1:52 am
    function debug_die() {
        call_user_func_array("debug", func_get_args());
        die;
    }
    
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