在C ++中将结构复制到数组中

Question

I've a struct of the following type

typedef struct Edge
{
    int first;
    int second;

}Edge;

Which I'm instantiating in my main function and copying into an array

Edge h_edges[NUM_EDGES];
for (int i = 0; i < NUM_VERTICES; ++i)
    {
        Edge* e = (Edge*)malloc(sizeof(Edge));
        e->first = (rand() % (NUM_VERTICES+1));
        e->second = (rand() % (NUM_VERTICES+1));
        memcpy(h_edges[i], e, sizeof(e));
    }

I keep running into the following error.

src/main.cu(28): error: no suitable conversion function from "Edge" to "void *" exists  

Line 28 is the line where the memcpy happens. Any help appreciated.

Answer 1

You don't need to use malloc or memcpy at all. You can just:

Edge h_edges[NUM_EDGES];
for (int i = 0; i < NUM_VERTICES; ++i)   // or should this be NUM_EDGES??
{
    h_edges[i].first = (rand() % (NUM_VERTICES+1));
    h_edges[i].second = (rand() % (NUM_VERTICES+1));
}

Answer 2

The first parameter of memcpy takes a pointer and the third argument needs to the size of the struct , not the pointer.

memcpy(&h_edges[i], e, sizeof(*e));

is the fix.

But this is ill-advised. Strictly the behaviour of your program is undefined . You can't copy a structure using memcpy in a defined way. Essentially the reason for this lies in structure packing.

Your best bet is to copy the structure members explicitly. You could build a function to do that.

An other point is that using % will introduce statistical bias in your random numbers, unless the modulus is a multiple of the generator's periodicity; which is unlikely.

(A small point, write Edge* e = malloc(sizeof(Edge)); instead. ie drop the cast on the right hand side. It's unnecessary in C.)