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I have a struct that has space for unsigned ints:

typedef struct {
    unsigned int *arr;
} Contents;

When I allocate memory: 

Contents *Allocator()
{
    Contents *cnt = malloc(sizeof(Contents));
    cnt->arr = calloc(1, sizeof(unsigned int));
}

I later retrieve it by dereferencing it by passing in a pointer to Contents and doing:

void SomeFunction(Contents *cnt)
{
   unsigned int * arr = cnt->arr;
   arr[0] >>= 1; // In the future 0 will be replaced by a loop over the array items
   cnt->arr = arr;
 }

Once I exit out of the function, cnt->arr becomes empty. Do I have to do a memcpy? Am I not understanding how the struct is laid out? As I understand 

cnt->arr = (*cnt).arr

Thanks!

Jens
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Rio
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  • Do you do `cnt->arr = arr` after `arr` was modified? What's the purpose of it? – Alexey Kukanov Apr 30 '11 at 06:05
  • Why not show the code of `SomeFunction()` (or code that exhibits the problem) along with the code that actually calls it. That way people don't have to guess at what's really happening when you say things like, "once I exit out of the function, `cnt->arr` becomes empty". – Michael Burr Apr 30 '11 at 06:30
  • That's correct. I do things like arr[0] |= 1 << 2; to arr (just as an example). – Rio Apr 30 '11 at 06:32
  • Please show the calling code - the `cnt->arr = arr` last line of `SomeFunction()` should have no effect (`cnt->arr` will remain unchanged from what it was at the start of `SomeFunction()`) since `arr` isn't changed after it's initialized. Please make sure the code you post demonstrates the problem (don't just post code that looks similar to what you're seeing the problem with). – Michael Burr Apr 30 '11 at 06:44

1 Answers1

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The problem is that you're doing unsigned int *arr = cnt->arr, which declares an unsigned int pointer and makes it point to cnt->arr. Once you modify the array, you then attempt to re-set the array - but by re-assigning pointers, you haven't changed the contents of the array; you've only changed the pointers. Thus, your cnt->arr = arr line doesn't actually change anything. Then, "unsigned int *arr" runs out of scope, and thus the pointer is destroyed, leaving you with unrecoverable data.

You'll need to copy the array somewhere temporary instead, and do your operations on that array instead, and then copy it back, OR (the easier method) just use your arr pointer and don't then try cnt->arr = arr - this effect will have been achieved anyway

Ben Stott
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  • So does this mean that if I do arr[0] >> 1 it will do the bit operations on cnt->arr if it is referenced by the pointer as above? – Rio Apr 30 '11 at 07:01
  • Yes, it does. Because they both point to the same location in memory, operating on one is the same as operating on the other. – Ben Stott Apr 30 '11 at 07:21
  • Assuming this answers the question, could I trouble you to mark this as solved? This way, others can see this when searching for a similar question! If it doesn't, what other information is needed? – Ben Stott Apr 30 '11 at 11:18