3

Take in mind the following piece of code:

#include <stdio.h>
#include <stdlib.h>

typedef struct
{
    int a;
    int b;
    int c;
}A;

A *test;

void init(A* a)
{
    a->a = 3;
    a->b = 2;
    a->c = 1;
}
int main()
{
    test = malloc(sizeof(A));
    init(test);
    printf("%d\n", test->a);
    return 0;
}

It runs fine! Now imagine that I want to use the malloc function outside the main itself without returning a pointer to the struct. I would put malloc inside init and pass test adress. But this doesnt seem to work.

#include <stdio.h>
#include <stdlib.h>

typedef struct
{
    int a;
    int b;
    int c;
}A;

A *test;

void init(A** a)
{
    *a = malloc(sizeof(A));
    *a->a = 3;
    *a->b = 2;
    *a->c = 1;
}
int main()
{
    init(&test);
    printf("%d\n", test->a);
    return 0;
}

It keeps telling me that int a(or b/c) is not a member of the struct A when I use the pointer.

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4
  • I bet that's not what it's telling you. What's the actual error message? Commented Jan 17, 2017 at 0:08
  • @melpomene request for member 'a' in something not a structure or union Commented Jan 17, 2017 at 0:10
  • 3
    See? It doesn't even mention struct A. It says the thing you're trying to use isn't a struct at all. (As the answers below explain, your code is getting parsed as *(a->a), a->a is the same as (*a).a, and the type of *a is A *; i.e. the compiler thinks you're trying to access members of a pointer, which makes no sense.) Commented Jan 17, 2017 at 0:13
  • and just an fyi,, there is no pass-by-reference in C,, it's all pass-by-value. In this case, you're passing a pointer by value to init where it's dereferenced and operated on. Commented Jan 17, 2017 at 0:22

4 Answers 4

Reset to default
6

Your problem is operator precedence. The -> operator has higher precedence than the * (dereference) operator, so *a->a is read as if it is *(a->a). Change *a->a to (*a)->a:

#include <stdio.h>
#include <stdlib.h>

typedef struct
{
    int a;
    int b;
    int c;
}A;

A *test;

void init(A** a)
{
    *a = malloc(sizeof(A));
    (*a)->a = 3;
    (*a)->b = 2;
    (*a)->c = 1;
}
int main()
{
    init(&test);
    printf("%d\n", test->a);
    return 0;
}
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Comments

5

You must add parenthesis:

void init(A **a)
{
    *a = malloc(sizeof(A)); // bad you don't verify the return of malloc
    (*a)->a = 3;
    (*a)->b = 2;
    (*a)->c = 1;
}

But it's good practice to do this:

void init(A **a)
{
    A *ret = malloc(sizeof *ret); // we want the size that is referenced by ret
    if (ret != NULL) { // you should check the return of malloc
        ret->a = 3;
        ret->b = 2;
        ret->c = 1;
    }
    *a = ret;
}
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Comments

4

You need to write (*a)->a = 3; for reasons of precedence.

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1 Comment

Wow didn't really think it was a question of syntax. Thank you!
4

Even though it's not a direct answer to your question, since we're in the vicinity of initialization I'd like to point out that C11 gives you a nicer syntax to initialize structs: 

void init(A **a)
{
    A *ret = malloc(sizeof *ret); // we want the size that is referenced by ret
    if (ret != NULL) { // you should check the return of malloc
        *ret = (A) {3, 2, 1};
        // or
        *ret = (A) { .a = 3, .b = 2, .c = 1 };
    }
    *a = ret;
}

Another advantage is that any uninitialized members are zeroed.

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1 Comment

It's very interesting an user friendly doc is here. But I want to know if it's create a temporary variable or if it's only affect the struct.

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