Question: As far as I can tell I've followed the advice given by other users, but I still get a segfault when I try to return a pointer to an array. The printf in main returns nothing, so I'm guessing that the pointer isn't to anything on the heap?

```
#include <stdio.h>
#include <stdlib.h>

int * stringy(int length){
   int *ptr = malloc(sizeof(int)*length);
   int i;
   for(i = 0; 1 < length; i++){
      ptr[i] = i;
   }
   return(ptr);
}

void main(int argc, char **argv){
   int strlen = 12;
   int *newptr = stringy(strlen);
   printf("%d\n",newptr[0]);
   free(newptr);
}

```

Answer: 
It seems like there is a typo in your code. Instead of

```
for(i = 0; 1 < length; i++){

```

, it looks like you meant

```
for(i = 0; i < length; i++){

```
You need to check for null pointers; `malloc` returns `NULL` when it can't allocate the memory you requested, and that would cause the problem you're referring to. Also, you should cast your pointer to `int *`.

The actual problem, however, is that you have a typo; rather than `1 < length`, your for loop condition should read `i < length`. Your for loop runs forever.