当为可为空的函数参数指定`null'时，如何使用默认值? [英] How to use default value when `null` is given for a nullable function parameter?

问题描述

在PHP 7.1中，调用以下函数时:

In PHP 7.1 when the following function is called:

private function doStuff(?int $limit = 999) { }

具有如下语法:

doStuff(null);

$ limit 的值变为 null .因此，我想可以说 $ limit 的值已显式设置为 null .

the value of $limit becomes null. So I guess it can be said that the value of $limit was explicitly set to null.

有什么办法可以克服这个问题?IE.当遇到空值(即缺少值)时，请使用默认值，是隐式还是显式?

Is there any way to overcome this? I.e. when a null value (i.e. the lack of a value) is encountered use the default, whether it is implicit or explicit?

推荐答案

没有PHP没有如果为null则默认还原"选项.您应该改为:

No PHP doesn't have a "fallback to default if null" option. You should instead do:

private function dostuff(?int $limit = null) {
    // pre-int typehinting I would have done is_numeric($limit) ? $limit : 999;
    $limit = $limit ?? 999;
}

或者，当您没有明智的做事价值时，请确保您执行 dostuff()或 dostuff(999).

Alternatively make sure you either do dostuff() or dostuff(999) when you don't have a sensible value for doing stuff.

注意:也有一些反思可以获取方法参数的默认值，但这似乎太多了.

Note: There's also reflection to get the default values of method parameters but that seems a too much.

但是，方法如下:

 $m = new ReflectionFunction('dostuff');
 $default = $m->getParameters()[0]->getDefaultValue();
 dostuff($default);

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