不能在c ++中打印指针的地址 [英] can not print address of a pointer in c++

问题描述

我是c ++的新手。我写了这个程序。

I'm new in c++. I wrote this program.

#include <iostream>

using namespace std;

int main(void) {
    int x=24;
    char y='A';
    char* pchar=&y;
    int* pint=&x;

    cout <<"pchar= "<<hex<<&pchar<<endl;
    cout <<"pint = "<<hex<<&pint<<endl;
    cout <<endl;
    cout <<"pchar= "<<hex<<pchar<<endl;
    cout <<"pint = "<<hex<<pint<<endl;

    pchar=NULL;
    pint=NULL;
    getchar();
    return 0;
}

及其结果

and its result

你能帮我理解为什么我不能打印没有& 的变量的地址？
我认为 pchar 已经是 y 的地址。

Can you help me to understand why I can't print the address of variables without &? I think pchar is already the address of y.

感谢

推荐答案

当您使用 char * （像你的变量 pchar ）使用运算符< <函数的字符串重载，并且当将其视为字符串时，它将打印字符，直到找到字符串终止符字符c $ c>'\0'）。不幸的是，在这种情况下， pchar 只指向一个字符，所以你有未定义的行为，因为函数出去寻找内存中的字符未分配

When you use a char* (like your variable pchar) the string overload of operator<< function is used, and when treating it as a string it will print characters until it find the string terminator character ('\0'). Unfortunately in this case pchar points only to a single character, so you have undefined behavior as the function goes out looking for characters in memory not allocated to you.

要打印指针，必须将其转换为 void * ：

To print a pointer you have to cast it to void*:

std::cout << "pchar = " << std::hex << reinterpret_cast<void*>(pchar) << '\n';

当您打印变量 pchar （当你打印& pchar ）时，因为表达式的类型是 char ** 它们没有直接重载，而是使用通用指针重载（此参考中的案例7） ）。

It works when you print the address of the variable pchar (when you print &pchar) because then the type of the expression is of type char** which have no direct overload, and instead uses the generic pointer overload (case 7 in this reference).

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