如何从结构中释放动态分配的内存？

Question

I haven't used C in a long while, and I'm still somewhat new to it. I'm confused by the syntax for pointers and references. Basically I have a struct container with pointers that I am dynamically allocating as arrays. I want to know how to free up that memory when I'm done with it. Here's what it looks like:

typedef struct {
    int* foo;
} Bar;

typedef Bar * BarRef;

BarRef newBar(int n) {
    BarRef B = malloc(sizeof(Bar));
    B->foo = calloc(n,sizeof(int));
}

/* This is what I am having trouble understanding */
void freeBar(BarRef *B) {
    free(B->foo);
    B->foo = NULL;
    free(B);
    *B = NULL;
}

I get a compiler error telling me that I am requesting a member from something that is not a struct. But I thought passing a Ref* would derefrence so It would be like passing the struct. I'm using gcc and ANSI C.

Answer 1

void freeBar(BarRef * B) {
    // Here B is a pointer to a pointer to a struct Bar. 
    // Since BarRef is Bar *
    // And B is a BarRef *
}

Since you want to modify a pointer to a struct Bar (setting the pointer to be NULL) you are correct in passing in a BarRef *, but the contents of your procedure should look like this:

free((*B)->foo);
(*B)->foo = NULL;
free(*B);
*B = NULL;

(*B)->foo works as follows:

(*B) dereferences B, given you a BarRef (AKA a Bar * ) (*B)->foo accesses the element called foo in the bar structure pointed to by (*B)

B->foo is invalid. It means access the element called foo in the bar structure pointed to by B . Since B doesn't point to a bar structure, but instead points to a pointer to a bar structure you get the "requesting a member from something that is not a struct" error. The "something that is not a struct" is a pointer to a struct.