无法执行变量php中返回的函数

I have a function with this syntaxe :

$variable = function() {
    return function(){
        //code
    }
}

I can't execute that function with

$variable()

Is there a way to execute this function ?

I don't know for what is it... But here example:

$a = function () {
   return function () { 
       return 1; 
   };
};
$b = $a();
echo $b();

The function is returning and the calling is not doing anything with it use echo:

echo $variable();

I would suggest having separate functions instead for clarity.

$variable = doSomething();

function doeSomething(){
   //code
}

Without assing to new variable you can do this:

call_user_func($variable());

Your anonymous function (outer function), stored in a $variable returns another function (inner function).

Calling $variable() means that outer function returns inner one. See - returns, not executes. But as you don't assign result of outer function to any variable, your inner function is not stored anywhere.

So, the solution is:

$res = $variable();  // res stores an inner function
$res();

Or (in PHP7)

$variable()();