在c程序中使用宏

Question

我用宏来搞定这个愚蠢的程序,但我不知道失败是什么:

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

#define READ_RX  (1 << 1)
#define WRITE_RX (1 << 2)
#define READ_TX  (1 << 3)
#define WRITE_TX (1 << 4)

#define READ_COMMAND(num) (num == 0) ? (READ_RX) : (READ_TX)
#define WRITE_COMMAND(num) (num == 0) ? (WRITE_RX) : (WRITE_TX)

int main(int argc, char **argv)
{

    printf("[DEBUG] 0x%04X\n", (READ_COMMAND(0)) | (WRITE_COMMAND(0))); //works fine
    printf("[DEBUG] 0x%04X\n", READ_COMMAND(0) | WRITE_COMMAND(0)); //doesn't work

    return 0;
}

结果:

$ ./test
[DEBUG] 0x0006 -> works fine
[DEBUG] 0x0002 -> doesn't work

有谁知道这是什么问题？

最好的祝福.

Answer 1

宏只是文本替换,意思是什么.即

(READ_COMMAND(0)) | (WRITE_COMMAND(0))

变

((num == 0) ? (READ_RX) : (READ_TX)) | ((num == 0) ? (READ_RX) : (READ_TX))

而

READ_COMMAND(0) | WRITE_COMMAND(0)

变

(num == 0) ? (READ_RX) : (READ_TX) | (num == 0) ? (READ_RX) : (READ_TX)

现在使用优先级规则,您可以看到,这是相同的

(num == 0) ? (READ_RX) : ( (READ_TX) | (num == 0) ? (READ_RX) : (READ_TX) )