C 程序 int 变量值在读取 Char 值后发生变化

Question

我是 C 编程新手，刚刚创建了一个小型计算器应用程序，但我注意到，当我在读取 Int 值后读取 char 值时，下一个直接 Int 变量会发生变化。到底是什么原因造成这种情况呢？这是我的代码

#include <stdio.h>

int main(){   
    int num1;
    int num2;
    char opr;
    int ans;

    printf("Enter the first number : ");
    scanf("%d", &num1);

    printf("Enter the second number : ");
    scanf("%d", &num2);

    printf("Enter the operater : ");
    scanf("%s", &opr);

    printf("%d \n", num1);
    printf("%d \n", num2);

    switch(opr) {
        case '+':
            ans=num1+num2;
            printf("The addtion of %d and %d is %d", num1, num2, ans);
            printf("\n");
            break;

        case '-':
            ans=num1-num2;
            printf("The substractuon of %d from %d is %d", num2, num1, ans);
            printf("\n");
            break;

        case '*':
            ans=num1*num2;
            printf("The multiplication of %d and %d is %d", num1, num2, ans);
            printf("\n");
            break;

        case '/':
            ans=num1/num2;
            printf("The substraction of %d from %d is %d", num1, num2, ans);
            printf("\n");
            break;
    }

    return 0;
}

Answer 1

这里你必须使用 scanf("%c", &opr); 而不是 scanf("%s", &opr); 因为 opr 是 char 你必须使用%c，其中 %s 用于扫描 String 。然后出现未处理的问题 '\n' 出现问题。因此，在 %c 前面添加一个额外的 '\n' 。因此该语句变为 scanf("\n%c", &opr);;

修改后的代码:-

#include <stdio.h>

int main()
{
    int num1;
    int num2;
    char opr;
    int ans;

    printf("Enter the first number : ");
    scanf("%d", &num1);

    printf("Enter the second number : ");
    scanf("%d", &num2);

    printf("Enter the operater : ");
    scanf("\n%c", &opr); // not scanf("%s", &opr);

    printf("%d \n", num1);
    printf("%d \n", num2);

    switch (opr)
    {
    case '+':
        ans = num1 + num2;
        printf("The addtion of %d and %d is %d", num1, num2, ans);
        printf("\n");
        break;

    case '-':
        ans = num1 - num2;
        printf("The substractuon of %d from %d is %d", num2, num1, ans);
        printf("\n");
        break;

    case '*':
        ans = num1 * num2;
        printf("The multiplication of %d and %d is %d", num1, num2, ans);
        printf("\n");
        break;

    case '/':
        ans = num1 / num2;
        printf("The substraction of %d from %d is %d", num1, num2, ans);
        printf("\n");
        break;
    }

    return 0;
}

输出:-

Enter the first number : 3
Enter the second number : 4
Enter the operater : *
3 
4 
The multiplication of 3 and 4 is 12