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​如何遍历复合文字数组以便打印book_idand 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("%d\n", i); 
        // how to iterate over compound literal array?
    }   
    return 0;
}
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1 回答 1

2

如何迭代复合文字数组？

你不能。

至少在不提供有关两个数组携带的元素数量（即 2 和 3）的附加信息的情况下并非如此。通过将两个数组分配给. 在运行时它不能再计算了。BookCodeMyBooks

您可以做的是定义一个标记值并将诸如停止器元素之类的实例添加到每个BookCode数组的末尾。这样每个数组的大小可以在运行时（重新）计算。

例如，这可以如下所示完成：

#include <stdio.h>
#include <string.h>

typedef struct
{
  int book_id;
  char value;
} BookCode;

#define BOOKCODE_STOPPER {-1, '\0'}
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("%zu\n", i);

    size_t s = get_codes_count(books + i);
    for (size_t j = 0; j < s; ++j)
    {
      printf("Book code %zu: id=%d, value=%c\n", 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书籍代码组合永远不会出现。在上面的示例中，我{-1, '\0'}为此选择了。

于 2015-10-03T20:25:30.730 回答