最烦琐的解析C ++ 11 [英] Most vexing parse C++11

问题描述

我很困惑关于 Y y {X {}}; 这行是什么，什么是它与最烦琐的解析的连接。赞赏简短的说明：

  #include< iostream> 
 
 struct X {
 X（）{std :: cout<< X; } 
}; 
 struct Y {
 Y（const X& x）{std :: cout< Y; } 
 void f（）{std :: cout<< F; } 
}; 
 int main（）{
 Y y {X {}}; 
 y.f（）; 
} 
  

解决方案

>

它创建了一个临时 X 通过调用默认构造函数初始化它，然后使用它来初始化 Y 变量，调用 const X& 转换

如果您尝试使用旧式初始化语法写这个语句

  Y y 
  

那么所谓的最讨厌的解析会将此解释为函数，而不是变量，声明：一个名为 y 的函数，返回类型为 Y 和单个参数，其类型为a a）函数返回 X 。

您可以添加多余的括号，函数声明：

  Y y（（X（）））; 
  

或者，因为C ++ 11，你可以使用括号初始化作为你的例子。 p>

I'm confused about Y y {X{}}; what exactly this line does and what is its connection to the most vexing parse. A brief explanation is appreciated:

#include <iostream>

struct X {
  X() { std::cout << "X"; }
};
struct Y {
  Y(const X &x) { std::cout << "Y"; }
  void f() { std::cout << "f"; }
};
int main() {
  Y y {X{}};
  y.f();
}

解决方案

what exactly this line does

It creates a temporary X, value-initialising it by calling the default constructor, and then uses that to initialise a Y variable, calling the const X& conversion constructor.

where is connection to Most vexing parse

If you were to try to write this using old-school initialisation syntax

Y y (X());

then the so-called "most vexing parse" would interpret this as a function, rather than a variable, declaration: a function called y, with return type Y and a single parameter, whose type is a (pointer to a) function returning X.

You could add extra parentheses, so that it can't be interpreted as a function declaration:

Y y ((X()));

or, since C++11, you can use brace-initialisation as your example does.

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