CODEWITHSUNDEEP
phpclosuresvariable-variables
aborted
aborted

Reputation: 4541

Passing variable variable into closure throws a parse error?

I just noticed that if you try to put a variable variable into a use list of a closure, it throws a parse error. Example code:

foreach ($array as $key => $item) {
    $$key = $item->something;
    $this->doSomething($key, function ($param) use ($item, $$key) {
        $param->foo($$key);
    });
}

The parse error is this:

Parse error: syntax error, unexpected '$', expecting '&' or variable (T_VARIABLE) in

Is there something I'm doing wrong here? Why wont it let me pass the variable variable?

If I store the variable variable's value in another variable, I can pass it normally through use, but that's not optimal for my actual case.

Upvotes: 0

Views: 104

Answers (1)

Anders Carstensen
Anders Carstensen

Reputation: 4194

You say you don't want to save the value in another variable. But what about saving a reference to the variable? That should be functionally equivalent to your code.

foreach ($array as $key => $item) {
    $$key = $item->something;
    $otherVar = &$$key;
    $this->doSomething($key, function ($param) use ($item, $otherVar) {
        $param->foo($otherVar);
    });
}

If this doesn't work in "your actual case", please refine the code example or explain why.

Upvotes: 1

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