Implementing the exponential function with basic arithmetic operations - function

Question

For the purpose of the exercise, I have to implement the exponential function with the most basic arithmetic operations. I came up with this, where x is the base and y the exponent:
function expAetB() {
product=1;
for (i=0; i<y; i++)
{
product=product*x;
}
return product;
};
However, there are more basic operations than product=product*x;. I should somehow be able to insert instead another for loop which multiply and pass the result, but I can't find a way to do it without falling into an infinite loop.

Answer 1

In the same way that exponentiation is repeated multiplication, so multiplication is simply repeated addition.
Simply create another function mulAetB which does that for you, and watch out for things like negative inputs.
You could go even one more level and define adding in terms of increment and decrement, but that may be overkill.
See, for example, the following program which uses the overkill method of addition:
#include <stdio.h>
static unsigned int add (unsigned int a, unsigned int b) {
unsigned int result = a;
while (b-- != 0) result++;
return result;
}
static unsigned int mul (unsigned int a, unsigned int b) {
unsigned int result = 0;
while (b-- != 0) result = add (result, a);
return result;
}
static unsigned int pwr (unsigned int a, unsigned int b) {
unsigned int result = 1;
while (b-- != 0) result = mul (result, a);
return result;
}
int main (void) {
int test[] = {0,5, 1,9, 2,4, 3,5, 7,2, -1}, *ip = test;
while (*ip != -1) {
printf ("%d + %d = %3d\n" , *ip, *(ip+1), add (*ip, *(ip+1)));
printf ("%d x %d = %3d\n" , *ip, *(ip+1), mul (*ip, *(ip+1)));
printf ("%d ^ %d = %3d\n\n", *ip, *(ip+1), pwr (*ip, *(ip+1)));
ip += 2;
}
return 0;
}
The output of this program shows that the calculations are correct:
0 + 5 = 5
0 x 5 = 0
0 ^ 5 = 0
1 + 9 = 10
1 x 9 = 9
1 ^ 9 = 1
2 + 4 = 6
2 x 4 = 8
2 ^ 4 = 16
3 + 5 = 8
3 x 5 = 15
3 ^ 5 = 243
7 + 2 = 9
7 x 2 = 14
7 ^ 2 = 49
If you really must have it in a single function, it's a simple matter of refactoring the function call to be inline:
static unsigned int pwr (unsigned int a, unsigned int b) {
unsigned int xres, xa, result = 1;
// Catch common cases, simplifies rest of function (a>1, b>0)
if (b == 0) return 1;
if (a == 0) return 0;
if (a == 1) return 1;
// Do power as repeated multiplication.
result = a;
while (--b != 0) {
// Do multiplication as repeated addition.
xres = result;
xa = a;
while (--xa != 0)
result = result + xres;
}
return result;
}