如何在变量中使用变量来定义php函数，然后回显它

Question

I am trying to use variable in variable function, then to do some calculations and later use that variable and print out answer:

    <?php

$a($b)=function() {

if ($b == 10 ) {
    return 10 ;
}else{
    return 20 ;

     }
}

$a(10);
echo $a(10);

?>

I am getting error:

Fatal error: Can't use function return value in write context

Answer 1

Here $a($b) means invoking a function instead of declaration. you should define $a and pass $b as a variable to the function.

Change this to:

$a($b)=function() {

This:

$a=function($b) {

Whole PHPcode Try this code snippet here

<?php

ini_set('display_errors', 1);

$a = function($b)
{

    if ($b == 10)
    {
        return 10;
    } else
    {
        return 20;
    }
};

echo $a(10);
?>