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您可以在结构初始化内部创建数组吗?

[英]Can you create an array inside of a struct initialization?

So I have a struct: 所以我有一个结构:

typedef struct Board {
    size_t size;
    char* board;
} Board;

I was wondering if it was possible to then do something like this, during initialization of the struct: 我想知道是否有可能在结构初始化期间执行类似的操作:

Board boardStruct = {
    solutionLength, 
    char emptyBoard[size]
};

Unfortunately, when I try to do it this way I get the compilation error: expected expression before 'char' 不幸的是,当我尝试以这种方式进行操作时,出现编译错误: expected expression before 'char'

Any ideas? 有任何想法吗? I'm trying to avoid declaring the array outside of the struct initialization, but if that is the only option I guess that is the route I will have to go with. 我试图避免在结构初始化之外声明数组,但是如果这是唯一的选择,我想那是我必须走的路线。

You can do something like that : 您可以这样做:

#include <stdlib.h>

typedef struct Board {
    size_t size;
    char* board;
} Board;

int main()
{
  const int solutionLength = 3; /* or #define solutionLength 3 */

  Board boardStruct = {
    solutionLength, 
    malloc(solutionLength)
  };

  return 0;
}

or closer to your proposal : 或更接近您的建议:

#include <stdlib.h>

typedef struct Board {
    size_t size;
    char* board;
} Board;


int main()
{
  const int solutionLength = 3; /* or #define solutionLength 3 */
  char emptyBoard[solutionLength];

  Board boardStruct = {
    solutionLength, 
    emptyBoard
  };

  return 0;
}

@bruno's solution will work. @bruno的解决方案将起作用。 Another thing you could try is to put the array within the Board struct. 您可以尝试的另一件事是将数组放入Board结构中。 Eg: 例如:

typedef struct Board {
    size_t size;
    char board[size];
} Board;

Upside: Avoids a malloc/free for each Board. 优势:避免为每个开发板分配内存/释放内存。
Downside: Board is larger, so it costs more to copy it. 缺点:木板较大,因此复制它的成本更高。 Also, you must know how big the board is before your program runs. 另外,在程序运行之前,您必须知道电路板的大小。

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