Array of struct pointers - overrides struct

Question

I'm learning C and encountered a problem with structs.

Let's assume I have the following struct:

typedef struct {
  int x;
} Structure;

int main (void) {
  Structure *structs[2];
  for(int i = 0; i < 2; i++) {
    Structure s = {i};
    structs[i] = &s;
  }
  for(int i = 0; i < 2; i++) {
    printf("%d\n", structs[i]->x);
  }

  return 1;
}

The output is:

1
1

I don't understand why the new struct is overring the old one.

It might be a stupid problem. But I don't get it.

Thanks!

Solved:

typedef struct {
  int x;
} Structure;

int main (void) {
  Structure *structs[2];
  for(int i = 0; i < 2; i++) {
    Structure *s = (Structure *)malloc(sizeof(Structure));
    s->x = i;
    structs[i] = s;
  }
  for(int i = 0; i < 2; i++) {
    printf("%d\n", structs[i]->x);
    free(structs[i]);
  }

  return 1;
}

Kerrek SB · Accepted Answer · 2013-05-14 18:28:46Z

4

The object s doesn't live beyond the scope of the first for loop. Storing its address is pointless, and dereferencing it is undefined behaviour.

answered May 14, 2013 at 18:28

Kerrek SB

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1 Comment

user2221323 Over a year ago

Thanks! I got it! I changed my code to using the heap as storage.

Mahesh · Accepted Answer · 2013-05-14 18:58:39Z

3

Code has undefined behavior. You are holding the reference of a local automatic variable.

for(int i = 0; i < 2; i++) {
    Structure s = {i};
    structs[i] = &s;

} // life time of s ends here

All bets are off since code has UB. So, it doesn't matter what output you get.

edited May 14, 2013 at 18:58

answered May 14, 2013 at 18:28

Mahesh

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5 Comments

user529758 Over a year ago

It's not that the variable is local - if it was static, no UB would be here. The correct (complete) terminology is "local automatic" variable.

Keith Thompson Over a year ago

@H2CO3: Actually, it's "object with block scope and automatic storage duration". I don't believe the standard uses the term "local" for this -- or "variable" for that matter.

user529758 Over a year ago

@KeithThompson Right, rather "block scope" - but I didn't imply "variable" was part of the terminology either (that's why it's ouside the quotation marks and the italics).

Mahesh Over a year ago

@H2CO3 Thanks for correcting me to use the correct terminology.

user529758 Over a year ago

@Mahesh Thank Keith Thompson for that - it turned out that I wasn't entirely correct either :)

K Scott Piel · Accepted Answer · 2013-05-14 18:29:16Z

1

The Structs s = {i}; only has scope within the for loop where you declared it. Once you leave that loop, it no longer exists, even though you still have a pointer to it. It's all undefined behavior after that.

answered May 14, 2013 at 18:29

K Scott Piel

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