將Switch語句轉換為if / then和if / else語句

Question

我想將此switch語句轉換為if / else和if / then語句。 我的Switch語句是：

char option;
printf("Choose your option : ");
scanf("%c",&option);

switch(option){
    case 'a':
    case 'A': a = 20 + 10 ;
        printf("Addition process result:%d",a);
        break;
    case 'b':
    case 'B': a = 20 - 10 ;
        printf("Subtraction process result:%d",a);
        break;
    case 'c':
    case 'C': a = 20 * 10 ;
        printf("Multiplication process result:%d",a);
        break;
    case 'd':
    case 'D': a = 20 + 10 ;
        printf("Division process result:%d",a);
        break;
    default:
        printf("Invalid option");
    }

Answer 1

您只需：

if(option == 'a' || option == 'A') {
    // do whatever
}
else if (option == 'b' || option == 'B') {
    // do whatever
}

...其他如果

那么對於“無效選項”，您只有一個else {}。 如果第一個if或隨后的else的任何if的評估結果為true，則將跳過所有其他if。

Answer 2

像這樣：

#include <stdio.h>

int main() {
    char option;
    printf("Choose your option : ");
    scanf("%c",&option);

    if (option == 'a' || option == 'A') {
        int a = 20 + 10;
        printf("Addition process result:%d", a);
    } else if (option == 'b' || option == 'A') {
        int a = 20 - 10;
        printf("Subtraction process result:%d", a);
    } else if (option == 'c' || option == 'C') {
        int a = 20 * 10;
        printf("Multiplication process result:%d", a);
    } else if (option == 'd' || option == 'D') {
        int a = 20 + 10;
        printf("Division process result:%d", a);
    } else {
        printf("Invalid option");
    }
    return 0;
}