为什么std :: nullptr_t在C ++中不能与std :: cout一起使用? [英] Why does not std::nullptr_t work with std::cout in C++?

问题描述

我了解了std::nullptr_t，它是空指针文字nullptr的类型.

I learned about std::nullptr_t that is the type of the null pointer literal, nullptr.

然后我制作了一个小程序:

Then I made small program :

#include <iostream>

int main() 
{
    std::nullptr_t n1; 
    std::cout<<n1<<endl;
    return 0;
} 

在这里，nullptr_t是数据类型，而n1是变量，我正在尝试打印变量的值.但是，编译器给出了一个错误:

Here, nullptr_t is data type and n1 is variable and I'm trying to print the value of variable. But, Compiler give an error:

prog.cpp: In function 'int main()':
prog.cpp:6:11: error: ambiguous overload for 'operator<<' (operand types are 'std::ostream {aka std::basic_ostream<char>}' and 'std::nullptr_t')
  std::cout<<n1<<endl;

为什么std::nullptr_t在C ++中不能与std::cout一起使用?我这是怎么了?

Why does not std::nullptr_t work with std::cout in C++? What am I wrong here?

推荐答案

operator<<输出流具有多种不同类型的指针的重载，但不是std::nullptr_t ¹ .这意味着编译器无法确定要调用的重载，因为 any 接受指针的重载同样好. (例如，它接受C样式的字符串的char const *以及接受输出原始指针值的void const *.)

operator<< for output streams has overloads for multiple different types of pointers, but not std::nullptr_t ¹. This means that the compiler cannot determine which overload to call, because any of the overloads accepting a pointer are equally good. (For example, it accepts char const * for C-style strings, and also void const *, which will output the raw pointer value.)

解决此问题的一种方法是定义自己的重载，强制使用void const *重载:

One option to fix this would be to define your own overload that forces the use of the void const * overload:

std::ostream & operator<<(std::ostream &s, std::nullptr_t) {
    return s << static_cast<void *>(nullptr);
}

或者让它做其他事情:

std::ostream & operator<<(std::ostream &s, std::nullptr_t) {
    return s << "nullptr";
}

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注意:

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Notes:

正如注释中指出的，
• ¹ ，C ++ 17中有一个接受std::nullptr_t的重载，因此如果您使用的是合格的C ++ 17，这将不再是一个问题.实施.
• endl需要std::资格-但无论如何都应该在这里使用'\n'. (std::endl仅是需要冲洗流的好主意.)

• ¹ As pointed out in the comments, there is an overload accepting std::nullptr_t in C++17, so this will cease to be an issue if you are using a conforming C++17 implementation.
• endl needs std::-qualification -- but you should use '\n' here anyway. (std::endl is only a good idea when you need the stream flushed.)

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