嵌套 if 中的范围歧义 [英] Scope ambiguity in nested if

问题描述

假设有一个 Foo 类，例如

Let's say one has a class Foo such as

class Foo {
public:
  void bar();
  operator bool() const { return true; }
};

然后就可以了

if(Foo foo = Foo())
{
  if(Foo foo = Foo())
  {
    foo.bar();
  }
}

现在我无法理解这里发生的作用域解析(我原以为重新声明 foo 会出现编译器错误).

Now I'm having trouble grasping the scope resolution going on here (I would've expected a compiler error for redeclaring foo).

我希望 foo.bar() 在第二个 foo 上执行(它的范围更近")，但我能保证它实际上是一个与第一个 foo 不同的对象吗?此外，它们是否在各自的 if 块的末尾独立放置(调用它们的析构函数)?

I expect foo.bar() to execute on the second foo (its scope is "closer") but am I garanteed that it's actually a different object than the first foo? Furthermore, are they each independently disposed (their destructor called) at the end of their respective if blocks?

推荐答案

C++ 很乐意让你声明一个同名的变量，只要它在嵌套的作用域内，所以没有歧义.

C++ is quite happy for you to declare a variable with the same name, as long as it is inside a nested scope so there is no ambiguity.

我希望 foo.bar() 在第二个 foo 上执行(它的作用域更近")

I expect foo.bar() to execute on the second foo (its scope is "closer")

你说得对

但是我能保证它实际上与第一个 foo 是不同的对象吗?

but am I garanteed that it's actually a different object than the first foo?

是的

此外，它们是否在各自的 if 块末尾独立放置(调用它们的析构函数)?

Furthermore, are they each independently disposed (their destructor called) at the end of their respective if blocks?

是的

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