如何迭代复合文字数组以便可以打印 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("%d\n", i);
// how to iterate over compound literal array?
}
return 0;
}
最佳答案
how to iterate over compound literal array?
你不能。
至少在不提供有关两个 BookCode
数组所携带的元素数量(即 2 和 3)的附加信息的情况下是这样。后面的信息丢失了 通过将两个数组分配给 MyBooks
的指针类型元素。在运行时不能再进行计算。
您可以定义一个哨兵值,并将一个实例(例如 stopper 元素)添加到每个 BookCode
数组的末尾。这样每个数组的大小就可以在运行时(重新)计算。
例如可以如下所示完成:
#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图书代码组合永远不会出现。在上面的示例中,我为此选择了 {-1, '\0'}
。
关于c - 如何迭代复合文字数组,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32926513/