c - 开关盒中的枚举

Question

这段代码有什么问题:

#define str(x) #x
#define xstr(x) str(x)


typedef unsigned char   uint8_t;   
typedef enum
{

         RED = 0x64,
         GREEN = 0x65,
       /* other enum values */
         BLUE = 0x87

} Format;

char buffer[50];

/* other code and variables */

/* somewhere later in code */     

myformat = RED;
/* later calling format function */

MapFormattToString(myformat,&buffer);


void MapFormattToString(uint8_t format,char *buffer)
{    
    printf("format = %x\n",format);  /*format printf has output 64 */
    switch(format)
    {
    case RED:
        sprintf(buffer,"%s\n", xstr(RED));
        break;
    case GREEN:
        sprintf(buffer,"%s\n", xstr(GREEN));
        break;
    case BLUE:
        sprintf(buffer,"%s\n", xstr(BLUE));
        break;
    default:
        sprintf(buffer,"Unsupported color\n");
    }
}

如果我使用 myformat = RED 单步执行此函数，它不会通过任何案例，而是通过切换案例中的默认值。
我的目标是缓冲区中应该有 RED 而不是相应的枚举值，即 64。

编译器:Windows XP 上的 gcc 3.4.5

Answer 1

我刚写了下面的程序，编译，测试，输出结果是:

$ ./test
    d
    e

$

这正是您所期望的。希望这可以帮助您发现程序中的一些差异。

#include<stdio.h>

typedef unsigned char uint8_t;

typedef enum {
    RED = 0x64,
    GREEN = 0x65,
    BLUE = 0x87
} Format;

void MapFormatToString(uint8_t format, char *buffer) {
    switch (format) {
        case RED:
            sprintf(buffer, "%c\n", RED);
            break;
        case GREEN:
            sprintf(buffer, "%c\n", GREEN);
            break;
        case BLUE:
            sprintf(buffer, "%c\n", BLUE);
            break;
        default:
            sprintf(buffer, "Unknown\n");
    }

}

main (int argc, char *argv[]) {
    char buffer[100];

    MapFormatToString(RED, buffer);
    printf(buffer);
    MapFormatToString(GREEN, buffer);
    printf(buffer);
    MapFormatToString(BLUE, buffer);
    printf(buffer);
}