reinterpret_cast问题 [英] reinterpret_cast question

问题描述

你能帮忙检查一下为什么代码没有正确打印变量charPtr1和charPtr2吗？

Could you help to check why the code didn't print the variable charPtr1 and charPtr2 correctly?

这是我的代码，谢谢

结果是：

地址= 00000032D70FFDA8; a = 100

地址= 00000032D70FFDAC; b = 101

地址= cd; c = c

地址 =; d =

address=00000032D70FFDA8;a=100
address=00000032D70FFDAC;b=101
address=cd;c=c
address=;d=

#include <iostream>
using namespace std;

struct sample
{
    public:
       int a;
       int b;
       char c;
  		 char d;
};


void  main(void) 
 {
  	sample s1;
  	s1.a=100;
  	s1.b=101;
  	s1.c='c';
  	s1.d='d';
 	sample *s=&s1;
 	  int *intPtr1=reinterpret_cast<int*>(s);
  	cout<<"address="<<intPtr1<<";a="<<*intPtr1<<endl;
  	
  	intPtr1++;
  	int *intPtr2=reinterpret_cast<int*>(intPtr1);
  	cout<<"address="<<intPtr2<<";b="<<*intPtr2<<endl;
  	
  	intPtr1++;
  	char *charPtr1=reinterpret_cast<char*>(intPtr1);
  	cout<<"address="<<charPtr1<<";c="<<*charPtr1<<endl;
  	
  	
  	intPtr1++;  	
  	char *charPtr2=reinterpret_cast<char*>(intPtr1);
  	cout<<"address="<<charPtr2<<";d="<<*charPtr2<<endl;
  	

}

推荐答案

On 3 / 26/2017 7:25 PM，cppBeginnnnnner写道：

On 3/26/2017 7:25 PM, cppBeginnnnnner wrote:

你能帮忙检查一下为什么代码没有正确打印变量charPtr1和charPtr2吗？ />


  &NBSP; char * charPtr1 = reinterpret_cast< char *>（intPtr1）; &NBSP; &NBSP; cout<<<" address ="<< charPtr1<<" ;; c ="<< * charPtr1<< endl;

Could you help to check why the code didn't print the variable charPtr1 and charPtr2 correctly?

    char *charPtr1=reinterpret_cast<char*>(intPtr1);     cout<<"address="<<charPtr1<<";c="<<*charPtr1<<endl;

运算符<<（char *）的重载，假设指针指向NUL终止的字符串，并打印该字符串。如果要打印地址，请将指针转换为void * first：

There's an overload of operator<<(char*), which assumes the pointer points to a NUL-terminated string, and prints that string. If you want to print the address, cast the pointer to void* first:

cout<<"address="<<(void*)charPtr1;

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