#include <stdio.h>

int main(void)
{
    unsigned long int number, num1, num2, mask;
    int i;
    char str[33];
   
    printf("Please input num1 and num2 : ");
    scanf("%lu %lu", &num1, &num2);
   
    /*********Found************/
    mask = __________________;
    number = num1 ^ num2;
    for(i=0; i<32; i++)
    {
        /*********Found************/
        str[i] = __________________;
        mask >>= 1;
    }
    str[32] = '\0';
    printf("%lu ^ %lu = %s\n", num1, num2, str);
   
    return 0;
}   

就是对每一位进行异操作啊

不大明白，这题怎么做？

#include <stdio.h>
#include <stdlib.h>

int main()
{
    unsigned long int number, num1, num2, mask;
    int i;
    char str[33];

    printf("Please input num1 and num2 : ");
    scanf("%lu %lu", &num1, &num2);

    /*********Found************/
    mask = 2147483648;
    number = num1 ^ num2;
    for(i=0; i<32; i++)
    {
        /*********Found************/
        str[i] = (number & mask) ? '1' : '0';
        mask >>= 1;
    }
    str[32] = '\0';
    printf("%lu ^ %lu = %s\n", num1, num2, str);

    return 0;
}

[此贴子已经被作者于2016-3-4 08:26编辑过]

此题目的难点在于如何将number按二进制顺序存入字符数组中。
设number=10=              00000000000000000000000000001010;
mask=pow(2,31)=2147483648=10000000000000000000000000000000;

第1次运算number&mask，i=0:
00000000000000000000000000001010
&
10000000000000000000000000000000
=
00000000000000000000000000000000(=0,false)
str[0]='0',mask >>= 1,mark=01000000000000000000000000000000

第2次运算number&mask，i=1:
00000000000000000000000000001010
&
01000000000000000000000000000000
=
00000000000000000000000000000000(=0,false)
str[1]='0',mask >>= 1,mark=00100000000000000000000000000000

......

第29次运算number&mask，i=28:
00000000000000000000000000001010
&
00000000000000000000000000001000
=
00000000000000000000000000001000(>0,true)
str[28]='1',mask >>= 1,mark=00000000000000000000000000000100

第30次运算number&mask，i=29:
00000000000000000000000000001010
&
00000000000000000000000000000100
=
00000000000000000000000000000000(=0,false)
str[29]='0',mask >>= 1,mark=00000000000000000000000000000010

第31次运算number&mask，i=30:
00000000000000000000000000001010
&
00000000000000000000000000000010
=
00000000000000000000000000000010(>0,true)
str[30]='1',mask >>= 1,mark=00000000000000000000000000000001

第32次运算number&mask，i=31:
00000000000000000000000000001010
&
00000000000000000000000000000001
=
00000000000000000000000000000000(=0,false)
str[31]='0',mask >>= 1,mark=00000000000000000000000000000000

最后str[33]="00000000000000000000000000001010"。综上，用1与number相对应的二进制位与运算，用结果的布尔值来判断对应位是0还是1，加入字符数组输出。

[此贴子已经被作者于2016-3-3 23:24编辑过]