How to define local functions in PHP?

Question

When I try to do this:

function a()
{
    function b() { }
}

a();
a();

I get Cannot redeclare b....
When I tried to do this:

function a()
{
    if(!isset(b))
        function b() { }
}

a();
a();

I get unexpected ), expected ....
How can I declare the function as local and to be forgotten of when a returns? I need the function to pass it to array_filter.

Answer 1

The idea of "local function" is also named "function inside function" or "nested function"... The clue was in this page, anonymous function, cited by @ceejayoz and @Nexerus, but, let's explain!

1. In PHP only exist global scope for function declarations;
2. The only alternative is to use another kind of function, the anonymous one.

Explaining by the examples:

1. the function b() in function a(){ function b() {} return b(); } is also global, so the declaration have the same effect as function a(){ return b(); } function b() {}.

2. To declare something like "a function b() in the scope of a()" the only alternative is to use not b() but a reference $b(). The syntax will be something like function a(){ $b = function() { }; return $b(); }

PS: is not possible in PHP to declare a "static reference", the anonymous function will be never static.

See also:

• var functionName = function() {} vs function functionName() {} (ans)
• What are PHP nested functions for?