#include <stdio.h> int main() { char m, n, op, result; scanf("%c%c%c", &m, &op, &n); if (m < '0' || m > '9' || n < '0' || n > '9') { printf("Invalid."); return 0; } m -= '0'; n -= '0'; switch ( ) { case '+' : result = m + n; break; case '-' : result = m - n; break; case '*' : result = m * n; break; case '/' : if (n != 0) {result = m / n; ;} default : printf("Invalid."); return 0; }
时间: 2024-01-20 11:03:14 浏览: 67
这段代码中缺少switch语句的case分支,导致程序无法正确计算输入的两个数字的四则运算结果。应该在switch语句中根据op的值进行相应的运算,并将结果赋值给变量result。同时,在除法运算中需要判断除数是否为0,以避免除以0的错误。修改后的代码如下:
```c
#include <stdio.h>
int main() {
char m, n, op, result;
scanf("%c%c%c", &m, &op, &n);
if (m < '0' || m > '9' || n < '0' || n > '9') {
printf("Invalid.");
return 0;
}
m -= '0';
n -= '0';
switch (op) {
case '+' :
result = m + n;
break;
case '-' :
result = m - n;
break;
case '*' :
result = m * n;
break;
case '/' :
if (n != 0) {
result = m / n;
} else {
printf("Invalid.");
return 0;
}
break;
default :
printf("Invalid.");
return 0;
}
printf("%d", result);
return 0;
}
```
这样就可以正确计算输入的两个数字的四则运算结果并输出了。
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