Optimizing string handling function
Const correctness

We don't modify the contents of str, so pass it as a const char*.  This allows us to give string literals as arguments without compiler warnings.

Check the arguments

What if str or substr is null?  What does it mean for length to be negative?  We should use size_t for indexing and lengths.

int overlap(const char *str, const char *substr, size_t length)
{
    if (!str || !substr)
        /* invalid arguments */
        return 0;


Add more tests

I'm not 100% sure from the description what the code is supposed to do.  That can be made clearer by providing more tests.  I recommend using a test framework to help identify the failures, but we can make our own for this:

int test(const char *a, const char *b, size_t len,
         int expected, const char *file, int line, const char *message)
{
    const int actual = overlap(a, b, len);
    if (actual == expected) return 0;

    fprintf(stderr, "%s:%d: %s should be %d, but got %d\n",
            file, line,
            message, expected, actual);
    return 1;
}

#define TEST(a, b, len, expected) \
    test(a, b, len, expected, __FILE__, __LINE__, "overlap(" #a ", " #b  ", " #len ")")

int main(void)
{
    return
        + TEST(NULL, NULL, 1, 0)
        + TEST("abc", "def", 0, 1)
        + TEST("abc", "abc", 4, 0)
        + TEST("abc", "abc", 3, 1)
        + TEST("abc", "bcd", 3, 0)
        + TEST("abc", "bcd", 2, 1)
        + TEST("bcd", "abc", 2, 1)
        + TEST("abcef01239", "abckef01798", 4, 1);
}


Algorithm

We're going more times over the data than we need to - although it looks like we're scaling as O(n²), it's actually O(n³), because of the if/else at the end of the loop that resets substr from time to time.

What we need to do is for each offset of str and substr, walk along the pair of them, counting character matches as we go (reset the count on a non-match).  If the count reaches length, then we've found a substring match.

As a worked example, let's test "ab1cdefg" against "bcde2":

ab1cdefg
bcde2
.....


In that first pass, nothing matched.  Now shift one of the strings by one character (i.e. index by n+1 instead of by n):

ab1cdefg
 bcde2
 1....


The longest run was 1, when we matched the b.  Next iteration:

ab1cdefg
  bcde2
  .123.


Here, we matched c, d and e successively, for a run length of 3.

In code, that looks something like:

#include &lt;string.h&gt;

/* internal method - returns true if a and b are equal for a run of
   length, without shifting */
static int overlap_run(const char *a, const char *b, size_t length)
{
    size_t count = 0;
    while (*a &amp;&amp; *b) {
        /* add one to count, or reset it to 0 */
        if (*a == *b)
            ++count;
        else
            count = 0;

        if (count == length)
            /* match found */
            return 1;

        ++a, ++b;
    }
    return 0;
}

int overlap(const char *a, const char *b, size_t length)
{
    if (!a || !b)
        /* invalid arguments */
        return 0;

    if (!length)
        /* every pair of strings has a zero-length match */
        return 1;

    const size_t la = strlen(a);
    const size_t lb = strlen(b);

    if (length &gt; la || length &gt; lb)
        /* strings are too short to match */
        return 0;

    for (size_t offset = 0;  offset &lt;= la - length;  ++offset)
        if (overlap_run(a+offset, b, length))
            return 1;

    for (size_t offset = 1;  offset &lt;= lb - length;  ++offset)
        if (overlap_run(a, b+offset, length))
            return 1;

    /* no match */
    return 0;
}


You might be able to avoid pipeline stalls by replacing

    if (*a == *b)
        ++count;
    else
        count = 0;


With

    count += (*a == *b);
    count *= (*a == *b);


But you'll need to benchmark to find out on your hardware.