为什么std :: cout将volatile指针转换为bool？ [英] Why does std::cout convert volatile pointers to bool?

问题描述

如果你试图找到一个volatile类型的指针，即使是一个volatile字符指针，你通常期望cout打印字符串，你会改为简单的得到'1'（假设指针不为null我想）。假设输出流运算符<<是专门为易失性指针的模板，但我的问题是，为什么？

If you try to cout a pointer to a volatile type, even a volatile char pointer where you would normally expect cout to print the string, you will instead simply get '1' (assuming the pointer is not null I think). I assume output stream operator<< is template specialized for volatile pointers, but my question is, why? What use case motivates this behavior?

示例代码：

#include <iostream>
#include <cstring>

int main()
{
    char x[500];
    std::strcpy(x, "Hello world");

    int y;
    int *z = &y;

    std::cout << x << std::endl;
    std::cout << (char volatile*)x << std::endl;

    std::cout << z << std::endl;
    std::cout << (int volatile*)z << std::endl;

    return 0;
}

输出：

Hello world
1
0x8046b6c
1

推荐答案

ostream :: operator<< 有以下重载：

ostream& operator<< (bool val );
ostream& operator<< (const void* val );

当你传递一个volatile指针时，第二个重载不能应用，因为volatile指针不能被转换到非易失性而没有显式铸造。但是，任何指针都可以转换为bool，所以选择第一个重载，你看到的结果是1或0.

When you pass in a volatile pointer, the second overload can't apply because volatile pointers cannot be converted to non-volatile without an explicit cast. However, any pointer can be converted to bool, so the first overload is chosen, and the result you see is 1 or 0.

所以真正的原因不是代表标准委托的有意决定，但简单地说，该标准没有指定一个带有易失性指针的重载。

So the real reason for this is not an intentional decision on behalf of the standards committe, but simply that the standard does not specify an overload that takes a volatile pointer.

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