c - C 编程中理解宏的问题

Question

如果我将“1”作为输入，以下代码的输出是什么？为什么？

仅供引用::iam 使用 gcc 构建代码。

#include <stdio.h>
int main() 
{
    // k integer to hold user input
    int k=0;

    scanf("%d",&k);

    switch(k)
    {
        case 1 :
        #define first_case
        break;
        case 2 :
         #define second_case
        break;
        case 3 :    
         #define third_case
        break;
        case 4 : 
        #define fourth_case
        break;
    }

#ifdeffirst_case printf("first_case\n"); #endif

#ifdef secondary_case printf("第二个案例\n"); #endif

#ifdef Third_case printf("第三个案例\n"); #endif

#ifdef Fourth_case printf("第四个案例\n"); #endif

}

Answer 1

宏 (#define MACRO ...) 在实际编译过程发生之前由 C 预处理器处理。

因此，只有在文件经过预处理后，编译器才会“看到”这一点:

int main()
{
  // your code goes here
  int k = 0;

  scanf("%d", &k);
  switch (k)
  {
  case 1:
    break;
  case 2:
    break;
  case 3:
    break;
  case 4:
    break;
  }

  printf("typed first_case\n");
  printf("typed second_case\n");
  printf("typed third_case\n");
  printf("typed fourth_case\n");
}

您可以这样编写程序，结果将完全相同:

#define first_case
#define second_case
#define third_case
#define fourth_case

/* Find the longest line among the giving inputs and print it */

#include <stdio.h>
int main()
{
  // your code goes here
  int k = 0;

  scanf("%d", &k);
  switch (k)
  {
  case 1:
    printf("case 1\n");
    break;
  case 2:
    break;
  case 3:
    break;
  case 4:
    break;
  }


#ifdef first_case
  printf("typed first_case\n");
#endif
#ifdef second_case
  printf("typed second_case\n");
#endif

#ifdef third_case
  printf("typed third_case\n");
#endif
#ifdef fourth_case
  printf("typed fourth_case\n");
#endif

}