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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