现在允许为std :: string分配一个数字？

问题描述

请考虑以下代码段：

  #include< iostream& 
 int main（）{
 std :: string foo; 
 foo = -1; //为什么编译器不会抱怨这个？ 
 std :: cout<< 1<< std :: endl; 
 std :: cout<< foo< std :: endl; 
 std :: cout<< 2< std :: endl; 
} 
  

实际输出（ideone.com C ++ 14模式和GCC 4.8。 4）：

<无输出> 

1. 为什么代码段完全编译？


2. 注释掉 foo = -1 ，我得到正确的stdout（ 1 和 2 ）。编译器使用 foo = -1; 编译会导致后续 cout 失败？

解决方案

  foo = -1; 
  

解析为 std :: string :: operator =（char） code>因为 -1 是 int 和 int 可以，在理论上，可以转换为 char 。

当 int 不表示有效的 char 时。

$ b

更新

C ++ 11标准（强调我）：

必须查阅您的编译器文档以了解它是否允许 char 对象保存负值，如果是，它如何处理这些对象。

Consider the following code snippet:

#include <iostream>
int main() {
    std::string foo;
    foo = -1; // why is the compiler not complaining about this?
    std::cout << "1" << std::endl;
    std::cout << foo << std::endl;
    std::cout << "2" << std::endl;
}

Actual output (both ideone.com C++14 mode and GCC 4.8.4):

<no output>

Questions:

1. Why did the code snippet compile at all?
2. Commenting out foo = -1, I get the correct stdout (1 and 2). What has the compiler compiled with foo = -1; that causes the subsequent couts to fail?

解决方案

foo = -1;

resolves to std::string::operator=(char) since -1 is an int and int can, in theory, be converted to a char.

It's not clear to me what the standard says when the int does not represent a valid char. It looks like in your implementation, the program crashes.

Update

From the C++11 Standard (emphasis mine):

It appears that you'll have to consult your compiler's documentation to understand whether it allows char object to hold negative values and, if it does, how does it treat such objects.

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