我有这个带有宏的愚蠢程序，但我不知道失败的原因是什么:
#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
有谁知道是什么问题吗？
最好的问候。
最佳答案
宏只是文本替换它们的意思。 即
(READ_COMMAND(0)) | (WRITE_COMMAND(0))
成为
((num == 0) ? (READ_RX) : (READ_TX)) | ((num == 0) ? (READ_RX) : (READ_TX))
鉴于
READ_COMMAND(0) | WRITE_COMMAND(0)
成为
(num == 0) ? (READ_RX) : (READ_TX) | (num == 0) ? (READ_RX) : (READ_TX)
现在使用优先规则，你可以看到，这和
(num == 0) ? (READ_RX) : ( (READ_TX) | (num == 0) ? (READ_RX) : (READ_TX) )
关于c - 在c程序中使用宏，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/42725786/