So when I pass a data type like a struct to assign some memory to it I find that the pointer doesn't change within the main scope. This further becomes a problem when I try to free the memory but obviously if its using the original pointer it will be pointing at the stack address.
void allocate(int *value){
value = malloc(10 * sizeof(int));
}
int main(){
int val2;
allocate(&val2);
free(&val2);
return 0;
}
I can fix this by using a double pointer to be passed into the allocate function but some course work I'm doing requires to only pass a pointer and I cant get it to update the pointer when it returns to main. I have looked around for a while but cant find a straight forward answer, I feel like my coursework is wrong but that might be my lack of understanding.
CodePudding user response:
The requirement to "only pass a pointer" seems contrived, and you could argue that a pointer to pointer (not a "double pointer") is a pointer, but perhaps you could use void *
to punch a hole in the type system. Or use a struct:
#include <stdlib.h>
#include <stdio.h>
struct intbuffer {
int *d;
size_t cap;
};
void *
xmalloc(size_t s)
{
void *r = malloc(s);
if( r == NULL ){
perror("malloc");
exit(1);
}
return r;
}
void
allocate(void *p, size_t s)
{
*(int **)p = xmalloc(s * sizeof(int));
}
void
allocate2(struct intbuffer *p)
{
p->d = xmalloc(p->cap * sizeof *p->d);
}
int
main(void){
int *val2;
struct intbuffer v;
allocate(&val2, 10);
free(val2);
v.cap = 10; /* Horrible api!! */
allocate2(&v);
free(v.d);
return 0;
}
Note that setting the capacity in the struct prior to making the call to allocate is a violation of many principles of software design, but this whole thing is absurdly contrived due to the bizarre artificial limitations.
CodePudding user response:
There are not enough *'s in each place, but you will have to figure out what that means.
void allocate(int** value){
*value = malloc(10 * sizeof(int));
}
int main(){
int* val2;
allocate(&val2);
free(val2);
return 0;
}