案例标签在C中没有减少到整数常量？

我正在开发一个游戏，我运行了我的代码并得到错误“案例标签不会减少到一个整数常量。” 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 #include 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？

Answer 1

您的一些案例标签是字符（类型为char ，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这些不是整数常量，不能以这种方式使用。

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

Answer 2

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':是的。

从概念上讲，您遇到了问题： jobs是一个int ，而您正在测试字符串。 如果你想允许用户输入字符串（多个单个字符），你需要保留一个字符串变量，并使用像fgets这样的东西来获得一整行输入。

Answer 3

你不能将字符串与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; } 

因为你只读取getch()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

1. 您的作业数组具有不一致的初始值设定项（mixed char和const char * ）

2. 您不能使用字符串文字作为案例标签，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; } 

3. 此外， player_cash是1个元素短（并且在索引[3]处写出了界限）

代码示例还展示了如何避免一般性gets不良，做一些基本的行端修剪并进行不区分大小写的比较（ stricmp的Windows，IIRC）： http ： stricmp

 #include #include #include 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; }