ipow.c

int64_t ipow(int64_t base, uint8_t exp) {
    static const uint8_t highest_bit_set[] = {
        0, 1, 2, 2, 3, 3, 3, 3,
        4, 4, 4, 4, 4, 4, 4, 4,
        5, 5, 5, 5, 5, 5, 5, 5,
        5, 5, 5, 5, 5, 5, 5, 5,
        6, 6, 6, 6, 6, 6, 6, 6,
        6, 6, 6, 6, 6, 6, 6, 6,
        6, 6, 6, 6, 6, 6, 6, 6,
        6, 6, 6, 6, 6, 6, 6, 255, // anything past 63 is a guaranteed overflow with base > 1
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
        255, 255, 255, 255, 255, 255, 255, 255,
    };

    int64_t result = 1;

    switch (highest_bit_set[exp]) {
    case 255: // we use 255 as an overflow marker and return 0 on overflow/underflow
        if (base == 1) {
            return 1;
        }
        
        if (base == -1) {
            return 1 - 2 * (exp & 1);
        }
        
        return 0;
    case 6:
        if (exp & 1) result *= base;
        exp >>= 1;
        base *= base;
    case 5:
        if (exp & 1) result *= base;
        exp >>= 1;
        base *= base;
    case 4:
        if (exp & 1) result *= base;
        exp >>= 1;
        base *= base;
    case 3:
        if (exp & 1) result *= base;
        exp >>= 1;
        base *= base;
    case 2:
        if (exp & 1) result *= base;
        exp >>= 1;
        base *= base;
    case 1:
        if (exp & 1) result *= base;
    default:
        return result;
    }
}

your code should end the cases sooner - when exp is down to { 0 , 1 , 2 }

Cuz from there it's very obvious what needs to be returned :::

   return { <= 0 := result
               1 := result * base
               2 := result * base * base  
          }

or more succinctly ::

  return ( base ** exp ) * result

Because when exp == 2 , you already know ahead of time that ( exp >>= 1 ) & 1 must be true, so why bother evaluating the obvious ?

Another speed up trick you can contemplate is that if you already know up-front that the exponent is 1 short of a power of 2, then you already know every single bit is a 1.

In that case, make it into a vanilla countdown loop for # of bits, and simply doing the exact same thing every round without bothering to either right shift or check for ( exp & 1 ). In my own code, for this scenario, I usually make the starting point slightly lagged, so each round I simply do a compound statement of

    result *= base *= base       # intentionally lagged, so one must be careful regarding when you 
                                 # need to perform the extra one to sync them back up with each other