X宏生成C风格字符串时意外含引号的解决求助

问题出在你对#NAME的使用上——#NAME本身会将宏参数转换为字符串字面量（比如#REG1会生成"REG1"），而你额外添加的\""会让最终的字符串包含引号字符。只需移除这些多余的引号即可解决问题。

修改以下两处代码：

1. 修正rnames数组定义
   原代码：

const char* const rnames[] = {
#define X(NAME,VALUE) (const char* const) "\"" #NAME "\"",
    rinfo
#undef X
};

改为：

const char* const rnames[] = {
#define X(NAME,VALUE) #NAME,
    rinfo
#undef X
};

（如果需要保留类型转换，可改为(const char* const) #NAME,，但这不是必须的）

2. 修正rmap的键值定义
   原代码：

std::map<std::string, RNAME> rmap = {
#define X(NAME,VALUE) { "\"" #NAME "\"", RNAME::RNAME_ ##NAME },
    rinfo
#undef X
};

改为：

std::map<std::string, RNAME> rmap = {
#define X(NAME,VALUE) { #NAME, RNAME::RNAME_ ##NAME },
    rinfo
#undef X
};

完整修正后的代码 #

#include <map>
#include <stdio.h>

#define rinfo \
  X( REG1, 10 )   \
  X( REG2, 20 )   \
  X( REG3, 33 )   \

enum class RNAME {
#define X(NAME,VALUE) RNAME_ ## NAME = __COUNTER__,
    rinfo
#undef X
};

const char* const rnames[] = {
#define X(NAME,VALUE) #NAME,
    rinfo
#undef X
};

std::map<std::string, RNAME> rmap = {
#define X(NAME,VALUE) { #NAME, RNAME::RNAME_ ##NAME },
    rinfo
#undef X
};

typedef struct {
    RNAME regname;
    const char* name;
    uint32_t value;
}regx_t;

regx_t regs[] = {
#define X(NAME,VALUE) { RNAME::RNAME_ ## NAME, rnames[ (int)RNAME::RNAME_ ## NAME ], VALUE },
    rinfo
#undef X
};

int main(int argc, char *argv[]){
    int nErrors = 0;
    int n = sizeof(rnames) / sizeof(const char*);

    for (int i = 0; i < n; i++) {
        std::map<std::string, RNAME>::iterator it;
        it = rmap.find(rnames[i]);
        if (it != rmap.end()) {
            int index = (int)it->second;
            printf("rname[%d] = %s = %u = %s\r\n", i, 
                        rnames[i], 
                        regs[index].value, 
                        regs[index].name);
        }
        else {
            ++nErrors;
        }
    }

    printf("RNAME::RNAME_REG1 = %d\n", RNAME::RNAME_REG1);
    printf("RNAME::RNAME_REG2 = %d\n", RNAME::RNAME_REG2);
    printf("RNAME::RNAME_REG3 = %d\n", RNAME::RNAME_REG3);

    printf("%d errors found\n", nErrors);

    return 0;
}

修改后的输出 #

rname[0] = REG1 = 10 = REG1
rname[1] = REG2 = 20 = REG2
rname[2] = REG3 = 33 = REG3
RNAME::RNAME_REG1 = 0
RNAME::RNAME_REG2 = 1
RNAME::RNAME_REG3 = 2
0 errors found

内容的提问来源于stack exchange，提问作者funbotix