C语言 如何迭代复合字面值数组

Question

如何遍历复合字面值数组，以便我可以打印book_id和value ?

#include <stdio.h>
#include <string.h>
typedef struct {
    int book_id;
    char value;
} BookCode;
typedef struct {
    BookCode *codes;
} Books;
int main() {
    Books MyBooks[] = { 
        (BookCode[]){ {1, 'a'},{2, 'b'} },
        (BookCode[]){ {1, 'd'},{2, 'c'}, {3, 'f'} },
    };  
    int i,j;
    int n1 = sizeof(MyBooks)/sizeof(MyBooks[0]);
    for(i = 0; i < n1; i++){
        printf("%dn", i); 
        // how to iterate over compound literal array?
    }   
    return 0;
}

Answer 1

如何迭代复合文字数组?

你不能。

至少不需要提供额外的关于两个BookCode数组所携带的元素数的信息，即2和3。通过将两个数组赋值给MyBooks的指针类型元素，后面的信息丢失了。在运行时不能再计算。

您可以做的是定义一个哨兵值，并在每个BookCode数组的末尾添加一个诸如stopper元素的实例。通过这种方式，可以在运行时(重新)计算每个数组的大小。

例如可以这样做:

#include <stdio.h>
#include <string.h>
typedef struct
{
  int book_id;
  char value;
} BookCode;
#define BOOKCODE_STOPPER {-1, ''}
static const BookCode BookCodeStopper = BOOKCODE_STOPPER;
typedef struct
{
  BookCode *codes;
} Books;
size_t get_codes_count(Books * books)
{
  BookCode * bc = books->codes;
  while (bc->book_id != BookCodeStopper.book_id
      && bc->value != BookCodeStopper.value)
  /* doing "while (memcmp(bc, &BookCodeStopper, sizeof BookCodeStopper)" might be faster. */
  {
    ++bc;
  }
  return bc - books->codes;
}
int main(void)
{
  Books books[] = {
    {(BookCode[]) {{1, 'a'}, {2, 'b'}, BOOKCODE_STOPPER}},
    {(BookCode[]) {{1, 'd'}, {2, 'c'}, {3, 'f'}, BOOKCODE_STOPPER}}
  };
  size_t n1 = sizeof books / sizeof books[0];
  for (size_t i = 0; i < n1; ++i)
  {
    printf("%zun", i);
    size_t s = get_codes_count(books + i);
    for (size_t j = 0; j < s; ++j)
    {
      printf("Book code %zu: id=%d, value=%cn", j, books[i].codes[j].book_id,
          books[i].codes[j].value);
    }
  }
  return 0;
}

这种方法意味着至少一种可能的图书代码组合永远不会出现。在上面的例子中，我选择了{-1, ''}