php - 为什么调用 __call 而不是 __callStatic

Question

我试图引用这个 question在 SO 上，但仍然不明白。

<?php

class A {
    public function __call($method, $parameters) {
        echo "I'm the __call() magic method".PHP_EOL;
    }

    public static function __callStatic($method, $parameters) {
        echo "I'm the __callStatic() magic method".PHP_EOL;
    }
}

class B extends A {
    public function bar() {
        A::foo();
    }

    public function foo() {
        parent::foo();
    }
}

(new B)->bar();
(new B)->foo();

据我了解，bar 函数静态调用类 A 上的 foo 方法，但 foo 方法使用 A 的实例调用该方法，该实例是 B 的父级。我期待它应该给我:

I'm the __callStatic() magic method
I'm the __call() magic method

但是，显然，我得到:

I'm the __call() magic method
I'm the __call() magic method

Answer 1

来自相关issue :

...A::foo() is not necessarily a static call. Namely, if foo() is not static and there is a compatible context ($this exists and its class is either the class of the target method or a subclass of it), an instance call will be made.

如果 foo() 是 static，它会按您预期的那样工作:

class A {
    public function __call($method, $parameters) {
        echo "I'm the __call() magic method $method".PHP_EOL;
    }

    public static function __callStatic($method, $parameters) {
        echo "I'm the __callStatic() magic method $method".PHP_EOL;
    }
}

class B extends A {
    public static function foo() { // <-- static method
        parent::foo();
    }
}

(new B)->foo();

I'm the __callStatic() magic method foo