Question

我正在开发游戏，我运行了我的代码并得到错误“案例标签不会减少到整数常量。”我想我知道这意味着什么，但我该如何解决呢？这是我的代码：

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

int player_cash[3] = {50};
char job[][20] {
    'A',
    'B',
    'C',
    "Donate",
    "Go to work",
    "Exit"
};
int jobs;

int main()
{
    while(player_cash[0] > 0) {
        printf("Please type A, B, C, Donate, Go to work, or Exit\n");
        switch(jobs) {

            case 'A':
            player_cash[0]-=5;
            player_cash[1]+=5;
            printf("Cash=%i\n\n", player_cash[0]);
            continue;

            case 'B':
            player_cash[0]-=5;
            player_cash[2]+=5;
            printf("Cash=%i\n\n", player_cash[0]);
            continue;

            case 'C':
            player_cash[0]-=5;
            player_cash[3]+=5;
            printf("Cash=%i\n\n", player_cash[0]);
            continue;

            case "Donate":
            player_cash[0]-=15; //Error here
            player_cash[1]+=5;
            player_cash[2]+=5;
            player_cash[3]+=5;
            printf("Cash donated\n\n");
            printf("Cash=%i\n\n", player_cash[0]);
            continue;

            case "Go to work":
            player_cash[0]+=10; //Error here
            printf("Work done\n\n");
            printf("Cash=%i\n\n", player_cash[0]);
            continue;

            case "Exit":
            printf("Thanks for playing!\n\n"); //Error here
            break;

            default:
            printf("Does not compute");
            continue;
        }
    }
        getchar();
        return 0;
}

因此，我希望用户做的是键入其中一个选项，并执行与其对应的操作。我该如何解决这个问题？

Answer 1

您不能将字符串用作case表达式：

case "Donate":

只能使用整数表达式，例如case 'A':没问题。

从概念上讲，您遇到了问题：jobs是int，您正在测试字符串。如果您想允许用户输入字符串（多个字符），您需要保留字符串变量，并使用类似fgets的内容来获得完整的输入。

Answer 2

您的一些案例标签是字符（char类型，用' s表示）。那些是整数常量。

其他标签是字符串文字（用"表示），其有效类型为const char *。¹那些不是整数常量不能以这种方式使用。

¹由于历史原因，它们通常可以像char *一样使用，但不要尝试更改它们。或者。

Answer 3

您无法将字符串与c进行比较。 "hello" == "hello"不会按预期工作。 switch只对基本类型进行简单的c比较。

switch(jobs) {

        case 'A':
        player_cash[0]-=5;
        player_cash[1]+=5;
        printf("Cash=%i\n\n", player_cash[0]);
        continue;

        case 'B':
        player_cash[0]-=5;
        player_cash[2]+=5;
        printf("Cash=%i\n\n", player_cash[0]);
        continue;

        case 'C':
        player_cash[0]-=5;
        player_cash[3]+=5;
        printf("Cash=%i\n\n", player_cash[0]);
        continue;

        case 'D':
        player_cash[0]-=15; //Error here
        player_cash[1]+=5;
        player_cash[2]+=5;
        player_cash[3]+=5;
        printf("Cash donated\n\n");
        printf("Cash=%i\n\n", player_cash[0]);
        continue;

        case 'G':
        player_cash[0]+=10; //Error here
        printf("Work done\n\n");
        printf("Cash=%i\n\n", player_cash[0]);
        continue;

        case 'E':
        printf("Thanks for playing!\n\n"); //Error here
        break;

        default:
        printf("Does not compute");
        continue;
    }

因为你只读了getch()中的一个字符，你可以比较这个值。（但要求用户只输入一个字母，因为他输入“Donate”，getch（）将首先读取D，返回，然后读取o等。）

Answer 4

您的作业数组初始化程序不一致（混合char和const char *）

您不能将字符串文字用作案例标签，因为char指针不是编译时常量。使用整数：

enum jobType
{
    jobA,
    jobB,
    jobC,
    jobDonate,
    jobGoToWork,
    jobExit,
    /* marker */
    jobInvalid
};

enum jobType typeOfJob(const char* const name)
{
    int i;
    for (i=jobA; i!=jobInvalid; ++i)
        if (0 == strcmp(jobNames[i], name))
            return i;
    return i;
}

此外，player_cash是1个元素短（并且在索引[3]处写出了界限）

代码示例还说明了如何避免一般gets不良，做一些基本的行尾修剪并进行不区分大小写的比较（stricmp在Windows，IIRC上）：http://liveworkspace.org/code/227015a4e51126d55ca4eb1eea739b02

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

int player_cash[4] = {50};

enum jobType
{
    jobA,
    jobB,
    jobC,
    jobDonate,
    jobGoToWork,
    jobExit,
    /* marker */
    jobInvalid
};

const char jobNames[][20] =
{
    "A",
    "B",
    "C",
    "Donate",
    "Go to work",
    "Exit"
};

enum jobType typeOfJob(const char* const name)
{
    int i;
    for (i=jobA; i!=jobInvalid; ++i)
#ifdef CASE_SENSITIVE
        if (0 == strcmp(jobNames[i], name))
#else
        if (0 == strcasecmp(jobNames[i], name))
#endif
            return i;
    return i;
}

const char* safer_gets()
{
    static char input[1024];
    char *p;
    const char* t;
    const char trimAt[] = "\r\n\t ";
    fgets(input, sizeof(input), stdin);

    for (t=trimAt; *t; ++t)
        while(p = strrchr(input, *t)) 
            *p = 0;

    return input;
}

int main()
{
    const char* input;
    while(player_cash[0] > 0)
    {
        printf("Please type A, B, C, Donate, Go to work, or Exit\n");
        input = safer_gets();

        switch(typeOfJob(input))
        {
        case jobA:
            player_cash[0]-=5;
            player_cash[1]+=5;
            printf("Cash=%i\n\n", player_cash[0]);
            continue;
        case jobB:
            player_cash[0]-=5;
            player_cash[2]+=5;
            printf("Cash=%i\n\n", player_cash[0]);
            continue;
        case jobC:
            player_cash[0]-=5;
            player_cash[3]+=5;
            printf("Cash=%i\n\n", player_cash[0]);
            continue;
        case jobDonate:
            player_cash[0]-=15; 
            player_cash[1]+=5;
            player_cash[2]+=5;
            player_cash[3]+=5;
            printf("Cash donated\n\n");
            printf("Cash=%i\n\n", player_cash[0]);
            continue;
        case jobGoToWork:
            player_cash[0]+=10;
            printf("Work done\n\n");
            printf("Cash=%i\n\n", player_cash[0]);
            continue;
        case jobExit:
            printf("Thanks for playing!\n\n");
            break;
        default:
            printf("Does not compute");
            continue;
        }
    }
    getchar();
    return 0;
}

案例标签在C中没有减少到整数常量？

4 个答案: