c - 在 C 程序中应用循环

Question

我已经坐下来编写一个非常简单的计算器，但我坚持应用给定的循环来返回给定的语句(如果它是错误的)。

如果用户输入错误的选项，我需要程序提示他们，直到输入正确的选项。

我该怎么做？

#include <stdio.h>

int main () {
    float num0, num1, result;
    int opt;



    printf("      ****************WELCOME TO ERIC'S CALCULATOR***************\n\n");

    printf("1-Addition\n");
    printf("2-Subtraction\n");
    printf("3-Multiplication\n");
    printf("4-Division\n");
    printf("5-Exit\n");
    printf("      *********???**************************************???******");
    printf("\n\n");

    printf("Choose An Option: ");
    scanf("%d", &opt);
    printf("\n");

    for(;;)

    if((opt != 1) || (opt != 2) || (opt != 3) || (opt != 4 ) || (opt !=5 ))     {       printf("Please Input Correct Option (1-5)\n");
        printf("Choose an Option: ");       scanf("%d", &opt);  }



    switch (opt){ case 1 :
    printf("                **********You Have Chosen Addition!**********\n\n");
    printf("Enter  First Number: ");
    scanf("%f", &num0);
    printf("Enter Second Number: ");
    scanf("%f", &num1);
              result = num0 + num1;
    printf("\n\n           %.1f + %.1f =  %.1f\n\n", num0, num1, result);
    break; case 2 :
    printf("                **********You Have Chosen Subtraction!**********\n\n");
    printf("Enter  First Number: ");
    scanf("%f", &num0);
    printf("Enter Second Number: ");
    scanf("%f", &num1);
              result = num0 - num1;
    printf("\n\n           %.1f - %.1f =  %.1f\n\n", num0, num1, result);
    break; case 3 :
    printf("                **********You Have Chosen Multiplication !**********\n\n");
    printf("Enter  First Number: ");
    scanf("%f", &num0);
    printf("Enter Second Number: ");
    scanf("%f", &num1);
              result = num0 * num1;
    printf("\n\n           %.1f * %.1f =  %.1f\n\n", num0, num1, result);
    break; case 4 :
    printf("                **********You Have Chosen Division!**********\n\n");
    printf("Enter  First Number: ");
    scanf("%f", &num0);
    printf("Enter Second Number: ");
    scanf("%f", &num1);
              result = num0 / num1;
    printf("\n\n           %.1f / %.1f =  %.1f\n\n", num0, num1, result);
    break; case 5 :
    printf("                **********THANK YOU FOR USING THIS SERVICE!**********\n\n");

    break;
    default:

        printf("Please Input Correct OPtion\n\n");
        break;


    }
    }

Answer 1

而不是

printf("Choose An Option: ");
scanf("%d", &opt);
printf("\n");

for(;;)

if((opt != 1) || (opt != 2) || (opt != 3) || (opt != 4 ) || (opt !=5 ))        
{       
    printf("Please Input Correct Option (1-5)\n");
    printf("Choose an Option: ");       scanf("%d", &opt);  

}

你可以这样做:

int inputFail = 1;  // Fake that input is wrong
while (inputFail)   // Keep on getting input as long as it is wrong
{
    printf("Choose An Option: ");
    scanf("%d", &opt);
    printf("\n");

    // Check if input is ok
    if((opt == 1) || (opt == 2) || (opt == 3) || (opt == 4 ) || (opt ==5 ))
    {
        // Mark input as ok (to end the while loop)
        inputFail = 0;
    }
}

或者另一种方式:

while (1) // Keep getting input "forever"
{
    printf("Choose An Option: ");
    scanf("%d", &opt);
    printf("\n");

    // Check if input is ok
    if((opt == 1) || (opt == 2) || (opt == 3) || (opt == 4 ) || (opt ==5 ))
    {
        // Input is ok so break out of the "forever" loop
        break;
    }
}

或更紧凑:

while (1)
{
    printf("Choose An Option: ");
    scanf("%d", &opt);
    printf("\n");
    if((opt >= 1) && (opt <= 5))
    {
        break;
    }
}