Passing a char * into char array

I have a function that returns a char *. I have another function that wants a char array as a parameter.

Code:

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char * func1() {
  char *temp = malloc(sizeof(char) * 10);
  strcpy(temp, "hello");
  return temp;
}

void func2(char temp[10]) {
  yadda yadda;
}

int main() {
  printf("%s", func2(func1()));
}

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Will that work as expected?

No. func2 does not return anything, so its non-existent return value cannot be used in main, because, well, it's non-existent.

bah! That was dumb...

Code:

---

char * func1() {
  char *temp = malloc(sizeof(char) * 10);
  strcpy(temp, "hello");
  return temp;
}

void func2(char temp[10]) {
  printf("%s", temp);
}

int main() {
  func2(func1());
}

---

how about that? I get a suspicious pointer conversion warning...

Quote:

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Originally Posted by Bladactania

how about that?

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You should store the return value of func1() so that you can use free() on it after you are done. Other than that yes, it should work (assuming that malloc() does not return a null pointer), but note that the char temp[10] parameter is effectively the same as char *temp, and that sizeof(char) is always equal to 1.

Quote:

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Originally Posted by Bladactania

I get a suspicious pointer conversion warning.

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Did you forget to include the three relevant headers, by any chance? It certainly looks like that in your example.

That should be legitimate; func2 expects a pointer to 10 characters, and you are passing it a pointer to 10 characters (and since the memory is malloc'ed, it's still a valid pointer). The compiler might not be able to figure that out, I guess.