php无法传递函数作为参数？ [重复]

Question

This question already has an answer here:

• Array_filter and empty() 4 answers

Works: returns [false,null]

array_filter([1, 2, 3, false, null], function($value) {
   return empty($value);
});

Does not Work:

array_filter([1, 2, 3, false, null], empty); // syntax error, unexpected ')',expecting '('

Why cant i just pass in the function as an argument?

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Answer 1

empty() is not a function, it's a language construct. array_filter expects a callable as the second argument. You will need to use a wrapper function exactly as you have in your first example:

array_filter([1, 2, 3, false, null], function($value) {
   return empty($value);
});