#include<stdio.h>int main(void){ char operator; double value1, value2; printf("Type in an expression: "); scanf("%lf%c%lf", &value1, &operator, &value2);switch(operator){ case '+': printf("=%.2f\n", value1+value2); break; case '-': printf("=%.2f\n", value1-value2); break; case '*': printf("=%.2f\n", value1*value2); break; case '/': if(value2==0) printf("Cannot divide into 0"); elseprintf("=%.2f\n", value1/value2); break; default: printf("Unknown operator\n"); break; } return 0;}哪里有问题

这段代码的逻辑没有问题，但是在除法运算中没有考虑除数为0的情况。如果除数为0，程序会输出"Cannot divide into 0"，但是并没有结束程序，会继续执行后面的语句。建议加上`return`语句来结束程序的执行。修改后的代码如下：

#include<stdio.h>
int main(void){
    char operator;
    double value1, value2;
    printf("Type in an expression: ");
    scanf("%lf%c%lf", &value1, &operator, &value2);
    switch(operator){
        case '+':
            printf("=%.2f\n", value1+value2);
            break;
        case '-':
            printf("=%.2f\n", value1-value2);
            break;
        case '*':
            printf("=%.2f\n", value1*value2);
            break;
        case '/':
            if(value2==0)
                printf("Cannot divide into 0");
            else
                printf("=%.2f\n", value1/value2);
            return 0; // 结束程序
        default:
            printf("Unknown operator\n");
            break;
    }
    return 0;
}