I simply want to write some code that makes use of recursion of functions to raise a base to its power. I know that recursion is not the most right way to do things in C++, but I just want to explore the concept a bit. The program asks the user for a base and an exponent and then console outs the answer. Here's the program I've written:
#include <iostream>
#include <math.h>
using namespace std;
int raisingTo(int, int);
int main()
{
int base, exponent;
cout << "Enter base value: ";
cin >> base;
cout << "Enter exponent value: ";
cin >> exponent;
int answer = raisingTo(base, exponent);
cout << "The answer is: " << answer << endl;
char response;
cin >> response;
return 0;
}
int raisingTo(int base, int exponent)
{
if (exponent > 0)
return 1;
else if (exponent = 0)
{
int answer = (int) pow((double)base, raisingTo(base, (exponent - 1)));
return answer;
}
}
The funny thing is, when I run this program, it keeps returning the answer as '1'! Can someone please help me on this?
To make this an actual C++ answer – this is the kind of task where you might consider making it a template function, as this should work with any kind of number type.
Recursion is in fact a good idea, but only if you make use of the benefits it can offer: it can avoid some of the multiplications, by factoring out low numbers from the exponent.
An example how many calculation this can save: suppose you want to raise a number to 19. That's 18 multiplications if you use the naïve loop-like approach. With this solution, what happens is
That was only 1+1+2+2 = 6 multiplications, 1/3 of the necessary amount for the loop-like approach! However, note that this doesn't necessarily mean the code will execute much faster, as the checking of factors also takes some time. In particular, the
%3
onunsigned
s is probably not faster than multiplication onint
s, so forNumT==int
it's not really clever at all. But it is clever for the more expensive floating point types,complex
, not to speak of linear algebra matrix types for which multiplication may be extremely expensive.Here's a cleaner explanation with O(log n) complexity
This algo works on following simple rules of exponent
Thus at each level, value of n is either half of what it was or it is little less than n . Thus the deppest the recursion will ever go is
1+log n
levelsInformation source
You have 3 main problems:
==
as=
is an assignment not compare.Your problem lies here
firstly, you've inverted the conditional (if the exponent is equal to zero, it should return), secondly, you are assigning and not comparing with the second
if
.Here's a version with better complexity (
O(lg exponent)
, instead ofO(exponent)
), which is conceptually similar to leftroundabout's version.It also uses tail recursion, which generally leads to better optimized machine code.