As my long title says: I am trying to return a pointer in c that has been dynamically allocated, I know, I have to free it, but I do not know how to myself, my search has showed that it can only be freed in main, but I cannot leave it up to the user to free the int.
My code looks like this right now,
int *toInt(BigInt *p)
{
int *integer = NULL;
integer = calloc(1, sizeof(int));
// do some stuff here to make integer become an int from a passed
// struct array of integers
return integer;
}
I've tried just making a temp variable and seeing the integer to that then freeing integer and returning the temp, but that hasn't worked. There must be a way to do this without freeing in main?
Your particular code gains nothing useful from dynamic allocation, as @unwind already observed. You can save yourself considerable trouble by just avoiding it.
In a more general sense, you should imagine that with each block of allocated memory is associated an implicit obligation to free. There is no physical or electronic representation of that obligation, but you can imagine it as a virtual chit associated at any given time with at most one copy of the pointer to the space during the lifetime of the allocation. You can transfer the obligation between copies of the pointer value at will. If the pointer value with the obligation is ever lost through going out of scope or being modified then you have a leak, at least in principle; if you free the space via a copy of the pointer that does not at that time hold the obligation to free, then you have a (possibly virtual) double free.
I know I have to free it, but I do not know how to myself
A function that allocates memory and returns a copy of the pointer to it without making any other copies, such as your example, should be assumed to associate the obligation to free with the returned pointer value. It cannot free the allocated space itself, because that space must remain allocated after the function returns (else the returned pointer is worse than useless). If the obligation to free were not transferred to the returned pointer then a (virtual) memory leak would occur when the function's local variables go out of scope at its end, leaving no extant copy of the pointer having obligation to free.
I cannot leave it up to the user to free the int.
If you mean you cannot leave it up to the caller, then you are mistaken. Of course you can leave it up to the caller. If in fact the function allocates space and returns a pointer to it as you describe, then it must transfer the obligation to free to the caller along with the returned copy of the pointer to the allocated space. That's exactly what the calloc() function does in the first place. Other functions do similar, such as POSIX's strdup().
Because there is no physical or electronic representation of obligation to free, it is essential that your functions document any such obligations placed on the caller.