for (unsigned i=0; i<len; i++)
        {
                if (s[i]=='o' && s[i+1]=='k')
                        return i; // found
        };

The last iteration of this loop, it accesses s[len-1] and s[len]. If the string is null terminated, that's OK. If it's really just a length qualified string, that's not OK.

The later implementations treat len differently.

EDIT: It's worse than I said. The later version won't detect "ok" at the end of the string, because they are off by one in the OPPOSITE direction.