Compilation error while accessing address of zero length array in C

Question

Possible Duplicate:
zero length arrays vs. pointers

Some new compilers throw up compilation error for below case

struct test {
int length;
char data[0];
};

int main(void)
{
char string[20] = {0};
struct test *t;

//Some code

memcpy(string, t->data, 19); //Compilation error 
}

However this gets resolved if I do like this.

memcpy(string, &(t->data[0]), 19);

Any reason why some new compilers are enforcing this restriction?

Edited to correct mistakes

Please consider if you think it would be helpful to include the actual error you're getting, verbatim from the compiler. — unwind, Nov 21 '12 at 09:52
This is error: error #170: pointer points outside of underlying object — nkumar, Nov 21 '12 at 10:18

score 6 · Answer 1 · edited May 23 '17 at 12:18

6

What's wrong with this:

struct test t;

memcpy(string, test->data, 19);

? Hint, test is a type.

EDIT: as to the real answer, see this question: zero length arrays vs. pointers (or similar questions on SO)

edited May 23 '17 at 12:18

Community

1
1

answered Nov 21 '12 at 09:54

Nim

33,299
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Sorry it should have been memcpy(string, t->data, 19); – nkumar Nov 21 '12 at 10:10

score 0 · Answer 2 · answered Nov 21 '12 at 10:01

0

Array can't have 0 size .

Here is the Standard:

ISO 9899:2011 6.7.6.2:

   If the expression is a constant expression, it shall have a value
   greater than zero

And second

use this:

memcpy(string,t->data,19); instead of what you have used.

answered Nov 21 '12 at 10:01

Omkant

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Yeah but this is in ISO c, the concept of zero lengh arrays to indicate a member which can latter be malloced is common. ISO C allows using array[]; for that (note no size given) where as gcc allows array[0]; – fkl Nov 21 '12 at 10:35

Compilation error while accessing address of zero length array in C

2 Answers2