“计算"和“字符地址" [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;

即使这样也没有.为什么?

Even this does not. Why?

推荐答案

我相信 << 运算符将其识别为字符串.将其转换为 void* 应该 工作:

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

cout << (void*)&p;

std::basic_ostream 有一个专门的运算符，它接受一个 std::basic_streambuf (基本上 is 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)

与接受任何指针的运算符相反(当然 char* 除外):

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

_Myt& operator<<(const void *_Val)

这篇关于“计算"和“字符地址"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！