C-将struct内部的指针重新分配给struct外部的指针

Question

I have two pointers, each pointing to a different array. One of the pointers is inside of a struct, as follows:

typedef struct
{
    int     N;    /* Number of Elements in array */
    double *c;    /* Pointer to an array         */
    /* Other members....  */
} CS;

The struct was initialized by:

CS->N = n; /* n is an integer of an initially unknown size */
CS->c = (double *)malloc(n * sizeof(double));

The pointer in the struct, CS->C , contains data that I no longer care about.

My other pointer was defined as follows:

double *alpha;
alpha = (double *)malloc(CS->N * sizeof(double));

I need to replace the contents of CS->C with alpha . I know I can do something naive like:

for (i=0;i<CS->N;i++) /* i is an integer */
    CS->c[i] = alpha[i];

I could also use memcpy, like this:

memcpy(CS->c,alpha,CS->N * sizeof(double));

My understanding is that both of these methods will copy the contents from memory located at alpha to the memory occupied by CS->C . That being a very expensive operation, it would make more sense to simply change the assignment of CS->C to alpha .

How can I do this?

I've tried to reassign the pointer by doing like CS->C = &alpha , but this gives me the following warning "assignment from incompatible pointer type".

Note: This is using ANSI C89 under full compliance, ie the compiler options are: -Wall -pedantic -ansi

Edit 1
Freeing CS->c and assigning it to alpha by doing:

free(CS->c);
CS->c = alpha;

does not work. It causes every entry in CS->c to become equal to 0.0 and it results in my program seg faulting.

Edit 2
I think I realized why the method suggested in my first edit did not work. alpha is a temporary pointer, created and initialized inside of a function, so once that function is exited, the memory occupied by alpha is "freed". Since CS->c points to that memory, it is also freed. Upon this discovery, I think I will rework my code, such that alpha and CS-c are initially swapped, such that when they are switched again, the end order will be correct. Thank you all for you valuable input.

Answer 1

Just copy the pointer

CS->C = alpha;

Alpha is a pointer to double, CS->C too, types matches, you simply change the pointer.

/!\\ Don't forgot to free the old CS->C ...

Answer 2

Free the old pointer and assing the new directly. (Not the adress of the new pointer)

The pointer in the struct, CS->c , contains data that I no longer care about.

My other pointer was defined as follows:

double *alpha;
alpha = malloc(CS->N * sizeof(double));

...

free (CS->c);
CS->c = alpha ; alpha=NULL;