如何从切换菜单返回

Question

我想制作一个菜单，将动物分类为昆虫、鸟类、哺乳动物和鱼类，每一种都有特定的特征（使用复合变量）。我想我会使用一个嵌套的开关盒（一个开关用于动物类别，另一个开关用于每个类别以执行操作）。问题是，例如，在我输入了一个新条目之后，我想可以选择返回并选择另一个操作，甚至返回并选择另一个类别。我怎样才能做到这一点？我尝试了 do-while，但我只是陷入了无限循环。

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
typedef struct{
    unsigned int noLegs, lifeSpan;
    char *name;
}insect;
typedef struct{
    float speed, length;
    char *habits, *name;
}bird;
typedef struct{
    float weight, height;
    char *food, *name;
}mammal;
typedef struct{
    float weight, depth, salt;
    char *name;
}fish;
int main(int argc, char **argv)
{
    char choice, action;
    bool running = true;
    bool returnBack = true;
    insect *i = (insect *)malloc(sizeof(insect));
    if(!i)
    {
        printf("Failed to allocate memory.\n");
        return -1;
    }
    while(running)
    {
        printf("Please choose a category:\n");
        printf("1. Insects\n2. Birds\n3. Mammals\n4. Fishes\n5. Exit\n");
        scanf(" %c", &choice);
        switch(choice)
        {
            case '1':
                printf("You've chosen: Insects\n");
                while(returnBack)
                {
                    printf("Please choose an action:\n");
                    printf("a. Add entry\nb. Delete entry\nc. Replace entry\nd. Lookup entries\ne. Back\n");
                    scanf(" %c", &action);
                    switch(action)
                    {
                        case 'a':
                            printf("What entry would you like to add?\n");
                            i->name = (char *)malloc(sizeof(char));
                            scanf("%s", i->name);
                            printf("How many legs does a %s have?\n", i->name);
                            scanf("%u", &(i->noLegs));
                            printf("What is the lifespan of a %s?\n", i->name);
                            scanf("%u", &(i->lifeSpan));
                            break;
                        case 'b':
                            printf("Which entry would you like to delete?\n");
                            break;
                        case 'c':
                            printf("Which entry would you like to replace?\n");
                            break;
                        case 'd':
                            printf("Which entry would you like to replace?\n");
                            break;
                        case 'e':
                            returnBack = false;
                            break;
                        default:
                            printf("Invalid action.\n");
                    }
                }
                break;
            case '2':
                printf("You've chosen: Birds\n");
                break;
            case '3':
                printf("You've chosen: Mammals\n");
                break;
            case '4':
                printf("You've chosen: Fishes\n");
                break;
            case '5':
                running = false;
                break;
            default:
                printf("Invalid choice.\n");
        }
    }
    return 0;
}

Answer 1

正如 alex01011 所说，你只需要 while 覆盖整个 switch
当用户选择 5 时，running 变为 FALSE 并退出循环。

不过有一件事，你忘了清除第一个 scanf()

中的空格

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

#define TRUE  (1==1)
#define FALSE (!TRUE)

/* ...
 * ALL THE STRUCTS
 */

int main(int argc, char **argv)
{
    char choice, action, running = TRUE;
    insect *i = (insect *)malloc(sizeof(insect));

    // CHECK ON MALLOC RETURN

    while(running)
    {
        printf("Please choose a category:\n");
        printf("1. Insects\n2. Birds\n3. Mammals\n4. Fishes\n5. Exit\n\n");
        scanf(" %c", &choice); // <-- The leading space was missing
        
    
        switch(choice)
        {
            /* ...
             * PREVIOUS CASES
             */
            case '5':
                running = FALSE;
                break;
                
            default:
                printf("Invalid choice.\n");
        }
    }

    return 0;
}

如何从切换菜单返回

How to go back from switch menu

c

switch-statement