Signed integer overflow

Question

Below in source code (I've commented the error on code), I get

Line 27: Char 33: runtime error: signed integer overflow: -1094795586+ -1094795586 cannot be represented in type 'int'

I've consulted different articles, like How to detected signed integer overflow but I'm not getting which integer to check.
Elements of working_array[] are filtered from nums[] if their value is less than target integer.
Since the elements of nums[] are like [ 3, 6, 11, 19, 2, ...] I'm not worried to checking for overflow during SUM on the last if statement. I'm wrong?

int* twoSum(int* nums, int numsSize, int target, int* returnSize){
    int count_elem = 0;
    int *working_array = (int*)malloc(numsSize * sizeof(int));
    
    // Going through original array
    for(int i=0;i<numsSize;i++){
        
        // If elem nums[i] is less than target, is accepted on a new working array
        if(nums[i] < target){
            count_elem += 1;
            working_array[count_elem] = nums[i];            
        }        
    }

    working_array = realloc(working_array, count_elem*sizeof(int));
    
    // Creating result array
    returnSize = sizeof(int)*2;
    int *result_array = (int*)malloc(returnSize);

    // Going through working array
    for(int i=0;i<count_elem;++i)
        for(int j=0;j<count_elem;++j)

            // SIGNED INTEGER OVERFLOW IN LINE BELOW
            if( (working_array[i]+working_array[j]) == target){
                result_array[0] = working_array[i];
                result_array[1] = working_array[j];
                free(working_array);
                return result_array;
            }
    
    free(working_array);
    return;
}

P.S: I know that cast malloc results is useless, but this is a minor issue maybe.

`returnSize = sizeof(int)*2;` : But `returnSize` is a pointer? You should get a big red warning on this. — Nate Eldredge, Nov 06 '21 at 00:04
Yes I was missing that. By using `sizeof(int)*2` in place of `returnSize` pointer, the issue persists the same... — bersi, Nov 06 '21 at 00:10
A lot of people don't believe that signed integer overflow is possible, or ask how to detect it. Yet it looks like your platform detected and rather nicely diagnosed it without your even asking! What platform is this? — Steve Summit, Nov 06 '21 at 00:46
@SteveSummit the error on this post, is related to LeetCode website's debugger. I was solving an exercise. (note that my question was about a compiling error, and not about how to solve the exercise itself) — bersi, Nov 08 '21 at 11:04
@bersi Fascinating. Thanks. That's a point in LeetCode's favor, then. — Steve Summit, Nov 08 '21 at 13:44

score 1 · Accepted Answer · answered Nov 06 '21 at 00:14

        if(nums[i] < target){
            count_elem += 1;
            working_array[count_elem] = nums[i];     
        }

Off-by-one error. This initializes working_array[1], working_array[2], ... potentially up to working_array[numsSize] which would be out of bounds. Meanwhile, working_array[0] is never initialized and contains garbage, which is probably the garbage value that provokes your overflow.

Make it:

        if(nums[i] < target){
            working_array[count_elem] = nums[i];     
            count_elem += 1;
        }

or perhaps

        if(nums[i] < target){
            working_array[count_elem++] = nums[i];     
        }

Also, as noted above,

    returnSize = sizeof(int)*2;
    int *result_array = (int*)malloc(returnSize);

is wrong because returnSize is a pointer. If the idea is to return the size by reference, then you want

    *returnSize = sizeof(int)*2;
    int *result_array = (int*)malloc(*returnSize);

Thank you for clarification. The first element of working_array, as initialized, solved the issue. Sorry again for missing the pointer used in the malloc... — bersi, Nov 06 '21 at 00:26

Signed integer overflow

1 Answers1