急!! C＋＋ 问题

[type="text"]

不知道怎么样display the memory address of the char!code:

#include<iostream>
#include<conio.h>
usingnamespace std;
int main()
{
char type_c;
cout<< "Address of type_c: "<<&type_c
<<" size is "<<sizeof(type_c)<<" bytes"<<endl;
return 0;
}
the code above doesn't work ! so any body can help me figure it out
thanks !

output should like this:
Address of char: XXXXXXXXX size is XX bytes

 

[苦逼热狗]

一定要用cout吗？printf不行吗？

#include<iostream>
int main()
{
char type_c;
printf("Address of type_c: %x, size is %d bytes\n", &type_c, sizeof(type_c));
return 0;
}
实在愿意用cout的话就
#include<iostream>
#include<conio.h>
usingnamespace std;
int main()
{
char type_c;
cout<< "Address of type_c: "<< hex <<(int)(&type_c)<<" size is "<<sizeof(type_c)<<" bytes"<<endl;
return 0;
}

 

[type="text"]

谢谢！！
不过！
printf("Address of type_c: %x, size is %d bytes\n", &type_c, sizeof(type_c));
这个里面的%x和%d是什么意思啊？？？
不懂！！
能解释一下吗？

 

[苦逼热狗]

printf - C++ Reference

C - printf and scanf, Format Specifiers

 

[type="text"]

thanks alot!!!!

 

[瓷娃娃]

你只是ＤＥＣＬＡＲＥ一个一　char 的ＶＡＲＩＡＢＬＥ type_c;并没有ＶＡＬＵＥ　ＡＳＳＩＧＮＭＥＮＴ，在ＤＥＣＬＡＲＥ的ＰＨＡＳＥ，，一个ＶＡＲＩＡＢＬＥ　是不会ＡＬＬＯＣＡＴＥ　ＭＥＭＯＲＹ的．．

 

[mooncake]

try:
cout << (void*)&type_c
you can also use "hex" to specify display an integer in hexadecimal or "dec" for normal decimal, such as
cout << hex << 1234 << dec << 1234;

A real c++ programmer should never use printf

 

[苦逼热狗]

productivity is my first concern.

 

[fmart.ca]

这个题目还真热闹啊。