为什么构造std :: string（0）不会发出编译器警告？ [英] Why does constructing std::string(0) not emit a compiler warning?

问题描述

说我有这段代码。

#include <string>

int main()
{
    std::string(0);
    return 0;
}

写 std :: string（0）会导致调用 std :: basic_string< char> :: basic_string（const char *），其中 0 作为此构造函数的参数，它将尝试将该参数视为指向C字符串的指针。

Writing std::string(0) results in std::basic_string<char>::basic_string(const char*) being called, with 0 as the argument to this constructor, which tries to treat the argument as a pointer to a C-string.

运行此代码显然会导致 std :: logic_error 被抛出。但是我的问题是：为什么GCC和MSVC 8.0都不会发出任何警告？我希望能看到从没有转换的整数中获取指针的思路。

Running this code obviously results in a std::logic_error being thrown. But my question is this : why both GCC and MSVC 8.0 don't emit any warnings? I'd expect to see something along the lines of "Making pointer from an integer without a cast".

推荐答案

0 是一个整数常量表达式，其值为0，因此它是一个空指针常量。将值0的常量用作空指针不是强制转换。

0 is an integer constant expression with value 0, so it is a null pointer constant. Using a 0-valued constant as a null pointer is not a cast.

C ++ 11引入了 nullptr （和 nullptr_t ），但是 0 作为空指针的处理在不久的将来不太可能改变代码取决于它。

C++11 introduces nullptr (and nullptr_t), but the treatment of 0 as a null pointer is unlikely to change in the near future as large amounts of code depends on it.

这篇关于为什么构造std :: string（0）不会发出编译器警告？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！