C语言 二进制 习题。。。求救。。输入两个二进制数，取反，与，或，异或。。显示结果。。

Question

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define M 32
char *extend(char *,char *);
char *reverse(char *,char *);
char *And(char *,char *,char *);
char *Or(char *,char *,char *);
char *orand(char *,char *,char *);
int main(int argc,char *argv[])
{
if(argc!=3)
{
puts("Wrong");
exit(1);
}
char x[M],y[M],z[M];
printf(" x=%s\n", extend(x,argv[1]));
printf(" y=%s\n", extend(y,argv[2]));
printf(" ~x=%s\n", reverse(z,x));
printf(" ~y=%s\n", reverse(z,y));
printf(" x&y=%s\n", And(z,x,y));
printf(" x|y=%s\n", Or(z,x,y));
printf(" x^y=%s\n", orand(z,x,y));
return 0;
}
char *extend(char *fd,char *fs)
{
int i;
for(i=0;i<=M-strlen(fs);i++)
fd[i]='0';
fd[i]='\0';
strcat(fd,fs);
return fd;
}
char *reverse(char *fd,char *fs)
{
char *save=fd;
strcpy(fd,fs);
while(*fd!='\0')
{
if(*fd=='0')
*fd='1';
else
*fd='0';
fd++;
}
return save;
}
char *And(char *fd,char *fs1,char *fs2)
{
char *save=fd;
while(*fs1!='\0')
{
if(*fs1=='1'&&*fs2=='1')
*fd='1';
else
*fd='0';
fs1++;
fs2++;
fd++;
}
return save;
}
char *Or(char *fd,char *fs1,char *fs2)
{
char *save=fd;
while(*fs1!='\0')
{
if(*fs1=='0'&&*fs2=='0')
*fd='0';
else
*fd='1';
fs1++;
fs2++;
fd++;
}
return save;
}
char *orand(char *fd,char *fs1,char *fs2)
{
char *save=fd;
while(*fs1!='\0')
{
if(*fs1!=*fs2)
*fd='1';
else
*fd='0';
fs1++;
fs2++;
fd++;
}
return save;
}

Answer 1

这个基本上已经写好了。
修改如下：
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define M 33
char *extend(char *,char *);
char *reverse(char *,char *);
char *And(char *,char *,char *);
char *Or(char *,char *,char *);
char *orand(char *,char *,char *);
int main(int argc,char *argv[])
{
char x[M],y[M],z[M];

if(argc!=3)
{
puts("Wrong");
exit(1);
}

printf(" x=%s\n", extend(x,argv[1]));
printf(" y=%s\n", extend(y,argv[2]));
printf(" ~x=%s\n", reverse(z,x));
printf(" ~y=%s\n", reverse(z,y));
printf(" x&y=%s\n", And(z,x,y));
printf(" x|y=%s\n", Or(z,x,y));
printf(" x^y=%s\n", orand(z,x,y));

return 0;
}
char *extend(char *fd,char *fs)
{
int i;
for(i=0;i<M-strlen(fs)-1;i++)
fd[i]='0';
fd[i]='\0';
strcat(fd,fs);
return fd;
}
char *reverse(char *fd,char *fs)
{
char *save=fd;
strcpy(fd,fs);
while(*fd!='\0')
{
if(*fd=='0')
*fd='1';
else
*fd='0';
fd++;
}
return save;
}
char *And(char *fd,char *fs1,char *fs2)
{
char *save=fd;
while(*fs1!='\0')
{
if(*fs1=='1'&&*fs2=='1')
*fd='1';
else
*fd='0';
fs1++;
fs2++;
fd++;
}
return save;
}
char *Or(char *fd,char *fs1,char *fs2)
{
char *save=fd;
while(*fs1!='\0')
{
if(*fs1=='0'&&*fs2=='0')
*fd='0';
else
*fd='1';
fs1++;
fs2++;
fd++;
}
return save;
}
char *orand(char *fd,char *fs1,char *fs2)
{
char *save=fd;
while(*fs1!='\0')
{
if(*fs1!=*fs2)
*fd='1';
else
*fd='0';
fs1++;
fs2++;
fd++;
}
return save;
}
你只需在DOS窗口的命令行输入即可：

如：执行程序文件名为OPT.exe
OPT 11101 10011
输出：
x=00000000000000000000000000011101
y=00000000000000000000000000010011
~x=11111111111111111111111111100010
~y=11111111111111111111111111101100
x&y=00000000000000000000000000010001
x|y=00000000000000000000000000011111
x^y=00000000000000000000000000001110