I can't solve this error : heap-buffer-overflow on Leetcode

Question

My code is below:

int* smallerNumbersThanCurrent(int* nums, int numsSize, int* returnSize){
    int count;
    int* out = malloc(500);
    int i, j;

    for(i = 0; i < numsSize; ++i){
        count = 0;
        for(j = 0; j < numsSize; ++j){
            if(nums[i] > nums[j]){
                count++;
            }
        }
        *(out + i) = count;
    } 

    return out;
}

and the error messages are below

Actually, this heap-buffer-overflow kept coming up when I've solved the problems on Leetcode. even if it works well in Terminal on Mac.

will appreciate any comments or helps on it!

Answer 1

Since I think this is homework I only describe what comes directly into my mind by looking at the code:

At first you allocate a 500 byte buffer on the heap ( malloc(500) ), at a point where you don't know the exact size yet. Also you don't check if malloc returned a NULL pointer. I would change this line to a calloc(numsSize, sizeof(*out)) , since the result array is at worst the same size as the input array. Also add a check for the return value.

The second thing is that you pass in a variable to hold the result count, but assign it nowhere. Without this information the caller can't know how many elements in the resulting array are valid.

Answer 2

At first, you don't check malloc function.

Secondly, the argument returnSize is not used in this function.

Your function will be failed if numsSize > (500/ sizeof(int))

May be the code below can help you

int* smallerNumbersThanCurrent(int* nums, int numsSize){
   int count;
   int* out = malloc(numsSize * sizeof(int));
   if (out) {
       fprintf(stderr, "%s: malloc failed\n", __func__);
       exit(EXIT_FAILURE);
   }

   for(int i = 0; i < numsSize; ++i){
       count = 0;
       for(int j = 0; j < numsSize; ++j){
            if(nums[i] > nums[j]){
               count++;
            }
       }
      *(out + i) = count;
  } 

   return out;
}