Question

我有一个菜单，其中一个选项是退出程序，但如果用户输入的字符不是1 2 3 4 5 6，它仍会退出程序或停止运行它。我希望在输入错误的字符后，菜单再次出现，用户可以再次输入。如果用户无限地输入错误的字符，我想要无限制地使用它。非常感谢！

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

int main(void)
{
  char opcao;


  printf("1 - \n"); 
  printf("2 -\n"); 
  printf("3 - \n"); 
  printf("4 - \n"); 
  printf("5 - \n"); 
  printf("6 - Terminar programa\n");  
  printf("Introduza a sua opcao:\n");
  scanf("%c",&opcao);

      switch(opcao){
    case'1':
      printf("Funcionalidade nao disponivel.");
      break;
    case'2':
      printf("Funcionalidade nao disponivel.");
      break;
    case'3':
      printf("Funcionalidade nao disponivel.");
      break;
    case'4':
      printf("Funcionalidade nao disponivel.");
      break;
    case'5':
      printf("Funcionalidade nao disponivel.");
      break;
    case'6':
      exit(0);
    default:
      printf("invalid input, please type again"); // this is what I want, but how?(now it would present the menu again...
      break;
      }

  return 0; 
}

Answer 1

使用do...while循环，以便您的代码如下所示：

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

int main(void)
{
  char opcao;


  printf("1 - \n"); 
  printf("2 -\n"); 
  printf("3 - \n"); 
  printf("4 - \n"); 
  printf("5 - \n"); 
  printf("6 - Terminar programa\n");  
  printf("Introduza a sua opcao:\n");
  do{ //loop
  scanf(" %c",&opcao); //discards blanks and reads the first non-whitespace character

      switch(opcao){
    case'1': 
    case'2':
    case'3':
    case'4':
    case'5':
  printf("Funcionalidade nao disponivel.");
   break;
    case'6':
      exit(0);
    default:
      printf("invalid input, please type again:"); // this is what I want, but how?(now it would present the menu again...
      }
  }while(opcao<'1' ||opcao>'6'); //loop until `opcao` less than '1' or greater than '6'

  return 0; 
}

Answer 2

我在这里给出了示例代码。您可以根据需要进行增强

#include <string.h>
int main (){
    char c, q=1;
    while ( q ){
        c=getchar ();
        switch (c){
            case '1':{} break;
            case '2': {printf ("quit the menu\n");q=0;}break;
        }
    }
    return 0;
}

如何在切换输入错误字符后重复菜单（再次请求用户输入）

2 个答案: