pow() function

[Ulutay]

hi, i have been trying to write the ready-made (cMath) pow function.

#include<iostream>
using namespace std;


float pow(float,int);

int main()
{
	cout<<pow(10,-2);
}

float pow(float a, int b)
{
	if(b==0)
	return 1;
	else if(b==1)
	return a;
	else if(b>1)
	return a*pow(a,b-1);
	

if(b==-1)
return 1/a;
if(b<-1)
{
a*=pow(a,b+1);
return 1/a;
}
	
	
}

I just cant figure out how to return 1/a ONCE in the base case instead of getting 1/a on each stack phase. i know what i wrote returns wrong value. if there was a better way to compute the number when powered by negative number please enlight me :)
Thanks

[D80DSL3A]

Just as you multiply by a for the case b>1 on line 19, you should divide by b for the b<1 case.

In any case I think you have over complicated the treatment. For example, you treat the case
b == 1 explicitly, even though one recursive step will take it to the b=0 case, which is handled already. You should need only the one terminating case.

I got this for the function myself. It seems to work:

float myPow( float b, int e )// using b=base and e=exponent = less confusion for me
{
    if( e == 0 ) return 1.0f;// terminates the recursion
    if( e > 0 ) return myPow( b, e-1 )*b;
    if( e < 0 ) return myPow( b, e+1 )/b;
}

EDIT: Note there is no need for the 'else if' because of the return encountered in the 'if' parts of the code. The return prevents the other if's from executing.

[dhayden]

Division is slow. so consider handling the negative exponent case like this:
if (e < 0) return myPow(1.0/b, -e);

Even with this change, the code will be slow for large exponents. You can speed it way up by noticing that b²⁷ = b¹⁶ * b⁸ * b² * b¹ and 27 = 16+8+2+1;

[Ulutay]

aha i got it now, thanks a lot, cheers !