“cout”和“char地址” [英] "cout" and "char address"

问题描述

char p;
cout << &p;

这不会打印字符p的地址。它打印一些字符。为什么？

This does not print the address of character p. It prints some characters. Why?

char p;
char *q;
q = &p;
cout << q;

即使这样也不会。为什么？

Even this does not. Why?

推荐答案

我相信<< 作为字符串。将其投射到 void * 应工作：

I believe the << operator recognizes it as a string. Casting it to a void* should work:

cout << (void*)&p;

std :: basic_ostream 它需要一个 std :: basic_streambuf （这基本上是一个字符串（在这种情况下））：

std::basic_ostream has a specialized operator that takes a std::basic_streambuf (which basically is a string (in this case)):

_Myt& operator<<(_Mysb *_Strbuf)

code> char * 当然）：

as opposed to the operator that takes any pointer (except char* of course):

_Myt& operator<<(const void *_Val)

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