不清楚最令人讨厌的解析 [英] Unclear most vexing parse

问题描述

让我们假设以下类为Foo。

  struct Foo 
 {
 int i; 
}; 
  

如果我想创建此类的实例并对其进行初始化，我应该这样做： Foo foo1 = Foo（）; 调用构造函数。

  int main（void ）
 {
 foo1 = Foo（）; 
 cout<< foo1.i<< ：<< foo1.j<<恩德尔// 0 
} 
  

然后我以为我会使用以下行，但不是：

  int main（void）
 {
 Foo foo2（ ）; //最令人讨厌的解析
 cout<< foo2<<恩德尔// 1 ???为什么？ 
 foo2.i ++; //错误：在'foo2'中请求成员'i'，该成员是非类类型'Foo（）'
} 
 
  

为什么foo2初始化为int（1）而不是Foo（）？

我最了解烦人的解析，它告诉我，根据我的理解，当一行可以解释为一个函数时，它将被解释为一个函数。

但是foo1（）被解释为

Foo foo2（）行之所以编译是因为它被解释为函数原型。好的。

为什么此行编译？ cout<< foo2<< endl;

是否打印foo2函数的地址？

正如您所说， Foo foo2（）; 是一个函数声明。对于 cout<< foo2 ， foo2 会衰减为函数指针，然后隐式转换为 bool 。作为非空函数指针，转换后的结果为 true 。

不使用 boolalpha ， std :: basic_ostream< CharT，Traits> :: operator<< 输出 1 true 。

如果boolalpha == 0，则将v转换为int类型，执行整数输出。

使用 std :: boolalpha 。

  cout<< boolalpha<< foo2<<恩德尔//打印 true 
  

Let's assume the following class Foo.

struct Foo
{
  int i;
};

if I want to make an instance of this class and initialize it, I should do: Foo foo1 = Foo(); to call the constructor.

int main(void)
{
    foo1 = Foo(); 
    cout << foo1.i << " : " << foo1.j << endl; // " 0 " 
}

then I thought I'd call the default constructor with the following line but it doesn't:

int main(void)
{
    Foo foo2(); // most vexing parse
    cout << foo2  << endl; // " 1 " ??? why?
    foo2.i++; // error: request for member ‘i’ in ‘foo2’, which is of non-class type ‘Foo()’ 
}

Why is foo2 initialized to int(1) and not Foo()?

I know about the most vexing parse, it tells that, from my understanding, when a line can be interpreted as a function, it is interpreted as a function.

But foo1() is interpreted as an int, not a function, nor a Foo class.

The line Foo foo2() compiles because it is interpreted as a function prototype. OK.

Why does this line compiles? cout << foo2 << endl;

Does it print the address of the foo2 function?

解决方案

As you said, Foo foo2(); is a function declaration. For cout << foo2, foo2 would decay to function pointer, and then converts to bool implicitly. As a non-null function pointer, the converted result is true.

Without using boolalpha, the std::basic_ostream<CharT,Traits>::operator<< would output 1 for true.

If boolalpha == 0, then converts v to type int and performs integer output.

You can see it more clearer with usage of std::boolalpha.

cout << boolalpha << foo2  << endl;  // print out "true"

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