使用C预处理获取字符串的整数值

Question

如何创建C宏来获取字符串的整数值？具体的用例来自这里的一个问题.我想改变这样的代码:

enum insn {
    sysenter = (uint64_t)'r' << 56 | (uint64_t)'e' << 48 |
               (uint64_t)'t' << 40 | (uint64_t)'n' << 32 |
               (uint64_t)'e' << 24 | (uint64_t)'s' << 16 |
               (uint64_t)'y' << 8  | (uint64_t)'s',
    mov = (uint64_t)'v' << 16 | (uint64_t)'o' << 8 |
          (uint64_t)'m'
};

对此:

enum insn {
    sysenter = INSN_TO_ENUM("sysenter"),
    mov      = INSN_TO_ENUM("mov")
};

在哪里INSN_TO_ENUM扩展到相同的代码.性能将是相同的,但可读性将大大提高.

我怀疑在这种形式下它可能是不可能的,因为C预处理器无法进行字符串处理,所以这也是一个不可取但可接受的解决方案(变量参数宏):

enum insn {
    sysenter = INSN_TO_ENUM('s','y','s','e','n','t','e','r'),
    mov      = INSN_TO_ENUM('m','o','v')
};

Answer 1

这是一个编译时的纯C解决方案，您指出可以接受。您可能需要将其扩展为更长的助记符。我将继续考虑所需的那个（即INSN_TO_ENUM("sysenter")）。有趣的问题：）

#include <stdio.h>

#define head(h, t...) h
#define tail(h, t...) t

#define A(n, c...) (((long long) (head(c))) << (n)) | B(n + 8, tail(c))
#define B(n, c...) (((long long) (head(c))) << (n)) | C(n + 8, tail(c))
#define C(n, c...) (((long long) (head(c))) << (n)) | D(n + 8, tail(c))
#define D(n, c...) (((long long) (head(c))) << (n)) | E(n + 8, tail(c))
#define E(n, c...) (((long long) (head(c))) << (n)) | F(n + 8, tail(c))
#define F(n, c...) (((long long) (head(c))) << (n)) | G(n + 8, tail(c))
#define G(n, c...) (((long long) (head(c))) << (n)) | H(n + 8, tail(c))
#define H(n, c...) (((long long) (head(c))) << (n)) /* extend here */

#define INSN_TO_ENUM(c...) A(0, c, 0, 0, 0, 0, 0, 0, 0)

enum insn {
    sysenter = INSN_TO_ENUM('s','y','s','e','n','t','e','r'),
    mov      = INSN_TO_ENUM('m','o','v')
};

int main()
{
    printf("sysenter = %llx\nmov = %x\n", sysenter, mov);
    return 0;
}