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