为什么"super"是关键字而不是Ruby中的方法? [英] Why is 'super' a keyword rather than a method in Ruby?

问题描述

在Ruby中， super 是关键字，而不是方法.

In Ruby, super is a keyword rather than a method.

为什么要这样设计?

Ruby的设计趋向于实现尽可能多的方法.关键字通常保留给具有自己语法规则的语言功能.但是，super的外观和行为类似于方法调用.

Ruby's design tends toward implementing as much as possible as methods; keywords are usually reserved for language features that have their own grammar rules. super, however, looks and acts like a method call.

(我知道在纯Ruby中实现super会很麻烦，因为它必须从caller或

(I know it would be cumbersome to implement super in pure Ruby, since it would have to parse the method name out of caller, or use a trace_func. This alone wouldn't prevent it from being a method, because plenty of Kernel's methods are not implemented in pure Ruby.)

推荐答案

它的行为略有不同，因为如果您不传递参数，则会传递所有当前参数(如果存在则为block). .我不确定该方法将如何工作.

It behaves a little differently, in that if you don't pass arguments, all of the current arguments (and block, if present) are passed along... I'm not sure how that would work as a method.

举一个比较人为的例子:

To give a rather contrived example:

class A
  def example(a, b, c)
    yield whatever(a, b) + c
  end
end

class B < A
  def example(a, b, c)
    super * 2
  end
end

我不需要处理产量，也不需要将参数传递给super.在您特别想传递不同参数的情况下，它的行为更像是方法调用.如果您根本不想传递任何参数，则必须传递空括号(super()).

I did not need to handle the yield, or pass the arguments to super. In the cases where you specifically want to pass different arguments, then it behaves more like a method call. If you want to pass no arguments at all, you must pass empty parentheses (super()).

它根本没有与方法调用完全相同的行为.

It simply doesn't have quite the same behaviour as a method call.

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