c - 无法将结构传递给结构内的指针

Question

给定以下定义:

#define arrayLengthInStruct 50

typedef struct {
   int _buf[arrayLengthInStruct];          
   int _bufLen;
} list;

// nested struct seems redundant but is there for some specific use
struct dev_req {   
    struct {    
        struct {       
            int src;
        }serviceReq;
    }_req;    
};

在主函数中

   int i=0;

// to pass the memory location
list g_src;                   // data is here
struct dev_req *dev_Areq;     // to be loaded/ transfer to here

   g_src._bufLen = arrayLengthInStruct;
// initialize the source
for (i = 0; i < g_src._bufLen; i++) {
    g_src._buf[i] = i+1;
    printf("%d \t", g_src._buf[i]);
}
printf("\n");

接下来的两行失败了

   //dev_Areq->_req.serviceReq.src = malloc(sizeof(g_src));    // failed
   //dev_Areq->_req.serviceReq.src = (list*) &g_src;           // failed

即使在通过前 1 行之后，下一部分是否可以工作也不清楚。

/*
for (i = 0; i < (dev_Areq->_req.serviceReq.src)->_bufLen; i++) {
    printf("%d \t", (dev_Areq->_req.serviceReq.src)->_buf[i]);
}
printf("\n");
*/

Answer 1

首先，dev_Areq是一个指针类型，它指向的内存还没有分配。

dev_Areq = (struct dev_req *)malloc(sizeof(struct dev_req ));

只有在那之后，您才能访问它的 _req 元素。

其次，你想将src指向一个list类型，并访问它的元素，那么你应该将src定义为list * 类型而不是 int。然后使用这一行将 src 指向 g_src:

dev_Areq->_req.serviceReq.src = (list*) &g_src;