c - 在用户不退出的情况下，在 C 中查找两个数字的总和、差等

Question

我需要找到用户输入的两个数字的总和、差异等。但是程序首先显示菜单，然后要求用户输入这些数字并进行操作。之后，它应该询问用户是否要再次计算一些东西，以及他们是否想要使用旧数字还是新数字。

我完全是 C 语言的初学者，所以我需要一些帮助。我在案例中遇到错误:

cfinal.c:49:14: warning: multi-character character constant [-Wmultichar]

还有这个:

cfinal.c:49:14: warning: overflow converting case value to switch condition type (21075 to 83) [-Wswitch]

有问题，我想不通。第二个问题是，我需要询问用户是否想再次重复，如果他们选择是，他们是想用新数字还是以前计算中使用的数字来做。那是我什至没有开始的事情，我不知道如何开始。谢谢。

我到目前为止的代码:

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>

int main (void)
{
    char choice;                //Define choice character type
    int first_number, second_number, firstSquare, secondSquare, firstQube, secondQube, cont;  //define intigers
    printf("Select operation you would like to do by typing shortcut:\n");
    printf("SU - sum\n");
    printf("DI - difference\n");
    printf("MU - multiplication\n");
    printf("RA - ratio\n");
    printf("SS - Sum of Squares\n");
    printf("SQ - Sum of Qubes\n");
    printf("RS - Sum of Squares\n");
    printf("RQ - Sum of Qubes\n");
    choice = (char)toupper(getchar());
    getchar ();
    printf("Enter the first number: ");
    scanf("%d", &first_number);
    getchar ();
    printf("Enter the second number:");
    scanf("%d", &second_number);
    getchar ();
    firstSquare = first_number * first_number;
    secondSquare = second_number * second_number;
    firstQube = first_number * first_number * first_number;
    secondQube = second_number * second_number * second_number;
    printf("\n");
    printf("Result is");
    switch(choice)
    {
        case 'SU':
            printf("%d + %d = %d", first_number, second_number, first_number + second_number);
            break;
        case 'DI':
            printf("%d - %d = %d", first_number, second_number, first_number - second_number);
            break;
        case 'MU':
            printf("%d * %d = %d", first_number, second_number, first_number * second_number);
            break;
        case 'SS':
            printf("%d * %d + %d * %d = %d", first_number, first_number, second_number, second_number);
            break;
        case 'SQ':
            printf("%d * %d * %d + %d * %d * %d = %d", first_number, first_number, first_number, second_number, second_number, second_number, firstQube + secondQube);
            break;
        case 'RS':
            printf("(%d * %d) / (%d * %d) = %d", first_number, first_number, second_number, second_number, firstSquare / secondSquare);
            break;
        case 'RQ':
            printf("(%d * %d * %d) / (%d * %d * %d) = %f", first_number, first_number, first_number, second_number, second_number, second_number, firstQube / secondQube);
            break;
        case 'RA':
            if (second_number == 0)
                fprintf(stderr, "The divisor can't be 0");
            else
            {
                printf("%d / %d = %f", first_number, second_number, (double)first_number / second_number);
            }
            break;
        default :
            fprintf(stderr, "Invalid entry");
            break;
    }
    printf("\n");
    for (;;)
    {
      printf("Continue (Type Y for yes or N for no)? ");
      cont = toupper(getchar());
      getchar ();
      if (cont == 'Y')
          return main(); // the difference.
      else if (cont == 'N')
          return EXIT_SUCCESS;
    }
}

Answer 1

getchar 返回单个字符。如果您想使用字符串“xy”、“d99”等，则必须读取一个字符串。 How to read a line from the console in C?

其次你不能打开你需要做的字符串

if(strcmp(input,"SU")==0)
{
....
}
else if (strcmp(input, "DQ") == 0)
{
}
...
else
{
... error