Here's an example indicating how we can use overflow from INT_MAX to INT_MIN to terminate a loop... ...if signed overflow would be properly defined.

    /* Write \"hi\" n times, counter will wrap to negative
     * to indicate we've written enough greetings.
     *
     * i=3  -> we want to write hi three times
     *
     * 1st INT_MAX-i+1 = INT_MAX-2 -> write hi
     * 2nd INT_MAX-1               -> write hi
     * 3rd INT_MAX                 -> write hi
     * 4th INT_MIN      (signed overflow, terminates loop)
     */

    int
    write_n_times_hi(int i)
    {
        int j;
        for (j=INT_MAX-i+1; j>0; j++)
                printf("hi\n");
        return i;
    }

This will be compiled into an endless loop by gcc 5.3.0/x86_64 for -O3, but will terminate properly with -O0.

If you change the (undefined for j=INT_MAX)

    j++

into the (defined)

    *((unsigned*)&j)++,

the compiler may no longer assume that j stays >0 all of the time, and the loop has to terminate.