Question

代码：

struct bunchofdata
{
    int i;
    void *dllist[i];
    int spltq[i];
    pthread_t tlist[i];
};

ERRORMSG：

error: ‘i’ undeclared here (not in a function)
  void *dllist[i];
               ^

我无法理解为什么这不起作用。帮助！

Answer 1

你可以这样使用，虽然你的结构的每个字段都是数组，但是这样的结构数组可能更好：

#include <malloc.h>
#include <pthread.h>

struct bunchofdata
{
    int i;
    void** dllist;
    int* spltq;
    pthread_t* tlist;
};

struct anotherbunchofdata
{
    int i;
    void* dll;
    int spltq;
    pthread_t tlist;
};

void init_bunchofdata( int size, struct bunchofdata* bd )
{
    bd->i = size;
    bd->dllist = malloc( size * sizeof( void* ) );
    bd->spltq = malloc( size * sizeof( int ) );
    bd->tlist = malloc( size * sizeof( pthread_t ) );
}

void free_bunchofdata( struct bunchofdata* bd )
{
    free( bd->dllist );
    free( bd->spltq );
    free( bd->tlist );
}



int main()
{
    struct bunchofdata bd;
    init_bunchofdata( 5, &bd );
    free_bunchofdata( &bd );

    return 0;
}

Answer 2

除了pikkewyn's answer之外，实现此目的的另一种方法是使用Flexible Array Member

这使得管理比分别分配成员更容易。

#include <stdio.h>
#include <stdlib.h>
typedef int pthread_t;
struct data
{
    void *dl;
    int spltq;
    pthread_t thread;
};
struct bunchofdata
{
    int i;
    struct data data_list[];
};
struct bunchofdata * data_factory(int size)
{
    struct bunchofdata * ret = malloc(sizeof(struct bunchofdata)
                                      +size*sizeof(struct data));
    /* fill in the members here*/
    ret->i=size;
    return ret;
}
int main(void) {
    struct bunchofdata *data10=data_factory(10);
    data10->data_list[9].spltq=0;
    printf("data10->data_list[9].spltq=%d",data10->data_list[9].spltq);
    free(data10)
    return 0;
}

得到“错误：在这里未声明（不在函数中）”与void * variable数组

2 个答案: