在函数C中将Struct复制到Pointer数组

Question

i have a huge problem allocating memory in C

i have this struct

typedef struct{
int x;
int y;
}T;

i want to create a function that dynamically adds a structs to a pointer. something like:

int main()
{
 T* t;
 f(&t);
 free(t);
}

up to this point i think everything is ok, now the function is where i get lost

void f(T** t)
{
 T t1;
 T t2;
 T t3;
 //first i malloc
 *t=malloc(sizeof(T)*T_MAX_SIZE);//i want another function to make the array bigger, but this is not as important as the problem
 t1.x=11;
 t1.y=12;
 t2.x=21;
 t2.y=22;
 t3.x=31;
 t3.y=32;
//now i want to copy the values from t1,t2,t3 to t[0],t[1],t[2]
 memcpy(&(*t[0]),&t1,sizeof(T));
 memcpy(&(*t[1]),&t2,sizeof(T));
 memcpy(&(*t[2]),&t3,sizeof(T));


}

i do not know the correct way of copying these structs.

the point of doing this is to use t out of the function (in the main)

many thanks :D

Answer 1

Your memcpy calls are incorrect.

In the expression &(*t[0]) , the array index has top precedence, followed by the pointer indirection. So with explicit parenthesis it looks like &(*(t[0])) .

So it first tries to array subscript t , which is the address of t in main. In the case of t[0] it still works, but t[1] references something past that variable, invoking undefined behavior. You want the array index of what t points to, which is (*t)[i] .

So the memcpy calls should be:

memcpy(&((*t)[0]),&t1,sizeof(T));
memcpy(&((*t)[1]),&t2,sizeof(T));
memcpy(&((*t)[2]),&t3,sizeof(T));

Answer 2

You don't need any copy functions to assign one structure to another - you simply equate them. So if you have

T var1 = {1, 2};
T var2 = var1;

the whole of var1 is copied to var2 . Amending your (simplified) program:

#include <stdio.h>
#include <stdlib.h>

#define T_MAX_SIZE 10

typedef struct{
    int x;
    int y;
}T;

void f(T** t)
{
    T t1;
    *t=malloc(sizeof(T)*T_MAX_SIZE);
    t1.x=11;
    t1.y=12;
    (*t)[0] = t1;
}

int main(void) {
    T* t;
    f(&t);
    printf ("Result %d %d\n", t[0].x, t[0].y);
    free(t);
    return 0;
}

Program output:

Result 11 12