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Home/ Questions/Q 7532319
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T05:25:03+00:00 2026-05-30T05:25:03+00:00

Consider the following example: // does not work foo( func_num_args() ); // works $args

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Consider the following example:

// does not work
foo( func_num_args() );

// works
$args = func_num_args();
foo( $args );

Why specifically does the former fail and the latter work?

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  1. Editorial Team
    Editorial Team
    2026-05-30T05:25:04+00:00Added an answer on May 30, 2026 at 5:25 am

    Note:

    Because this function depends on the current scope to determine
    parameter details, it cannot be used as a function parameter in
    versions prior to 5.3.0. If this value must be passed, the results
    should be assigned to a variable, and that variable should be passed.

    http://php.net/func_num_args

    In other words: because of the way this function works internally. The runtime is in a different state when it’s inside a regular function body and when it’s evaluating expressions to be used as parameters to another function. This was changed in 5.3.

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