Return makes integer from pointer without a cast (simple for loop)

Question

Why does this C code return the warning in the title?

char n_zeroes(int n) {
  char str[n];

  int i;
  for (i = 0; i < n; i++) {
    str[i] = '0';
  }

  return str;
}

Also see [Can a local variable's memory be accessed outside its scope?](http://stackoverflow.com/questions/6441218/can-a-local-variables-memory-be-accessed-outside-its-scope) — Lundin, Feb 09 '16 at 13:45

Iharob Al Asimi · Accepted Answer · 2016-02-09T13:25:14.070

4

Your function is declared to return char and you are returning an array of char. But even if you fix the return value which should be char * it's still wrong because you should not return a local variable, instead pass the array as an argument to the function like this

void
n_zeroes(int n, char *str)
{
    for (int i = 0 ; i < n ; ++i)
        str[i] = '0';
}

but this is essentially the same as

memset(str, '0', n);

also, note that this array filled just with zeroes does not represent a string, to make it a valid string that you can use with str* functions or pass to *printf() functions with the "%s" specifier it needs one extra element at the end with the specific value of '\0' or 0.

As you can understand from the error your code is making a pointer, from an integer without an explicit cast. This is because you are allowed to do that, i.e. to cast an integer to a pointer. Whether that's a correct thing to do or not is entirely a different thing, but the standard allows many things that can be misused too. That's the great thing about c, with great power comes greate responsibility.

edited Feb 09 '16 at 13:25

answered Feb 09 '16 at 13:06

Iharob Al Asimi

52,653
6
59
97

How do I declare the function to return an array? – Fela Maslen Feb 09 '16 at 13:06
1

You can't return an array you can return a pointer to it, and in this case that would be wrong. – Iharob Al Asimi Feb 09 '16 at 13:08
You can return a pointer to an array, but you would have to dynamically allocate and free it later using malloc and free. – Straw1239 Feb 09 '16 at 13:09
3

If the intention is to return the content of your local variable str, you better don't. Please read about using arrays and pointers in C before continuing, or your project is probably going to be a desaster... – Murphy Feb 09 '16 at 13:09
@iharob Feel free to flag my comment for deletion if you don't like my typos :-\ – Murphy Feb 09 '16 at 13:18
@Murphy What? I was trying to help you fix it, but never mind! – Iharob Al Asimi Feb 09 '16 at 13:20
@iharob Would have been too late anyway, the edit button was already gone (probably due to the votes?). Never mind, too. – Murphy Feb 09 '16 at 13:30
@Murphy No, it's because you can edit comments only within 5 minutes after posting them. – Iharob Al Asimi Feb 09 '16 at 13:31

Murphy · Answer 2 · 2016-02-09T18:51:38.983

0

The answer to your question is, because str is a pointer to an array, the compiler casts it to the intended return value (char, which translates to a signed integer ~~8 bits~~ usually 1 byte wide) and senses that there's something wrong between your code and your intentions...

The answer to your problem is, you can't do what you're intending here this way in C. To fill a char array with zeroes (a task better left to memset() instead of using a for loop) in a function and use it in the calling code, you need to pass a pointer to the reserved array to the function.

edited Feb 09 '16 at 18:51

answered Feb 09 '16 at 13:16

Murphy

3,827
4
21
35

Not necessarily `8` bits, `CHAR_BIT` bits. – Iharob Al Asimi Feb 09 '16 at 13:22
Well, as this is technically correct, I intend to postpone confusing the OP with details until later. For most platforms this approximation should do, and here it doesn't help to get the code right. – Murphy Feb 09 '16 at 13:28

Return makes integer from pointer without a cast (simple for loop)

2 Answers2