c语言 - 如何修复"return 0 / exit(0) not working properly inside for loop"？

Question

我正在制作一个科学计算器。并想到循环语句，直到用户使用 0 作为选项之一要求停止。但即使在输入 0 之后。它最后一次问这个陈述： printf("Enter two numbers (For only one no. required you can just enter other number anything)n");

我尝试使用 goto, exit(0) 和 return 0 语句。甚至while(1)和for(;;)循环。

#include <stdio.h>
#include<math.h>
int main()
{
    int a;
    float b,c;
    float d=3.14159/180;
    while(1)
    {
    printf("nScientific Calculator :n");
    printf("Enter option:n 0- Exit, 1-Add, 2-Sub, 3-Multiply, 4-Divide,n 5-sin(x), 6-cos(x), 7-tan(x), 8-sinh(x), 9-cosh(x), 10-tanh(x),n11-log10(x),12-exponent,13-power of x w.r.t y n");
    scanf("%d",&a);
    printf("Enter two numbers (For only one no. required you can just enter other number anything)n"); //Here is where it starts even after return 0
    scanf("%f%f",&b,&c); //Here after inputting value it ends.
    switch(a)
    {
        case 0:return 0; //Here is the return 0;
        case 1:printf("%d",(int)(b+c)); break;
        case 2:printf("%d",(int)(b-c)); break;
        case 3:printf("%d",(int)(b*c)); break;
        case 4:printf("%f",b/c); break;
        case 5:printf("%f",sin(b*d)); break;
        case 6:printf("%f",cos(b*d)); break;
        case 7:printf("%f",tan(b*d)); break;
        case 8:printf("%f",sinh(b*d)); break;
        case 9:printf("%f",cosh(b*d)); break;
        case 10:printf("%f",cosh(b*d)); break;
        case 11:printf("%f",tanh(b*d)); break;
        case 12:printf("%f",log10(b)); break;
        case 13:printf("%f",exp(b)); break;
        case 14:printf("%f",pow(b,c)); break;
        default:printf("Enter correct optionn");
    }
    }
    return 0;
}

我希望它只是退出并退出程序，但它printf("Enter two numbers ---n");要求输入，输入值后退出。

Answer 1

但即使在输入 0 之后。它最后一次问这个声明：printf("Enter two numbers (For only one no. required you can just enter other number anything)n");

您需要立即检查大小写"0"，因此

 printf("Enter option:n 0- Exit, 1-Add, 2-Sub, 3-Multiply, 4-Divide,n 5-sin(x), 6-cos(x), 7-tan(x), 8-sinh(x), 9-cosh(x), 10-tanh(x),n11-log10(x),12-exponent,13-power of x w.r.t y n");
 scanf("%d",&a);
 printf("Enter two numbers (For only one no. required you can just enter other number anything)n"); //Here is where it starts even after return 0
 scanf("%f%f",&b,&c); //Here after inputting value it ends.
 switch(a)
    case 0:return 0; //Here is the return 0;
    case 1:printf("%d",(int)(b+c)); break;
    ...

必须是

printf("Enter option:n 0- Exit, 1-Add, 2-Sub, 3-Multiply, 4-Divide,n 5-sin(x), 6-cos(x), 7-tan(x), 8-sinh(x), 9-cosh(x), 10-tanh(x),n11-log10(x),12-exponent,13-power of x w.r.t y n");
scanf("%d",&a);
if (a == 0)
  return 0;
printf("Enter two numbers (For only one no. required you can just enter other number anything)n"); //Here is where it starts even after return 0
scanf("%f%f",&b,&c); //Here after inputting value it ends.
switch(a)
   case 1:printf("%d",(int)(b+c)); break;
   ...

请注意，检查

输入的有效 int 也是一个好主意，因此要检查scanf("%d",&a)是否返回 1，对于scanf("%f%f",&b,&c)也是如此，因此检查它返回 2 ...

正如@OznOg在对情况 12 和 13 的评论中所说，您只需要读取一个数字，而不是两个。另请注意，如果代码无效，询问数字是没有用的，因此：

  if (scanf("%d",&a) != 1)
    return -1;
  if (a == 0)
    return 0;
  if ((a < 0) || (a > 14))
    printf("Enter correct optionn");
  else if ((a == 12) || (a == 13)) {
    printf("Enter one numbern");
    if (scanf("%f",&b) != 1)
      return -1;
    if (a == 12)
      printf("%f",log10(b)); 
    else
      printf("%f",exp(b)); break;
  }
  else {
    printf("Enter two numbersn");
    if (scanf("%f%f",&b,&c) != 2)
      return -1;
    switch(a)
    {
    case 1:printf("%d",(int)(b+c)); break;
    case 2:printf("%d",(int)(b-c)); break;
    case 3:printf("%d",(int)(b*c)); break;
    case 4:printf("%f",b/c); break;
    case 5:printf("%f",sin(b*d)); break;
    case 6:printf("%f",cos(b*d)); break;
    case 7:printf("%f",tan(b*d)); break;
    case 8:printf("%f",sinh(b*d)); break;
    case 9:printf("%f",cosh(b*d)); break;
    case 10:printf("%f",cosh(b*d)); break;
    case 11:printf("%f",tanh(b*d)); break;
    case 14:printf("%f",pow(b,c)); break;
    }
  }