hi there, can someone help me how to write a recursive function that takes the number to the power. Thus, if the number is 2 and the power is 4, the function will return 16.
I'm looking forward for anyone to help me with this my problem.
I try to solve it to myself but i can't find the right code for this.
Thanks...
Post what you have tried.
Post what you have tried.
This is my program i created.... please help to debug this
#include <iostream.h>
#include <stdlib.h>
int byThePower(int number, int power);
int main()
{
int number, power, answer;
cout <<"Enter a number: ";
cin >>number;
cout <<"\nEnter a Power: ";
cin >>power;
answer = byThePower(number,power);
cout <<"\n\nAnswer: "<<answer;
cout <<"\n\n";
system("PAUSE");
return 0;
}
int byThePower(int number, int power)
{
int ans,num,power1 ;
num =1;
if (power != 0)
{
number=number*number;
power--;
return(byThePower(number,power));
}
else
{
return number;
}
}Thanks
int ans,num,power1 ;
num =1;Why you declare something that you didn't use?
number=number*number;
power--;
return(byThePower(number,power));Let say that the number is 2 and the power is 4 then with this code it would process like this.
yeah that was the output of my program.. but i want my program to output like this:
2^4 = 16 or
3^3 = 27.
I don't know what is the right code to achieve this... can you help me to give the right code and explain why....
I hope this is not too much for you but im very thankful if you could help me..
>>3^3 = 27.
that's the wrong answer, so you program will never get it (unless the program is wrong) :)
you over-complicated your program -- the solution is much simpler
int byThePower(int number, int power)
{
if(power >1)
number = number * byThePower(number,power-1);
return number;
}I would write it like this: (this code works only when the power is a positive number)
int byThePower(int number, int power)
{
if (power >= 1)
return number * (byThePower(number , power - 1));
else
return 1;
}I would write it like this: (this code works only when the power is a positive number)
You are right -- the solution I posted doesn't work for power of 0.
I would write it like this: (this code works only when the power is a positive number)
int byThePower(int number, int power) { if (power >= 1) return number * (byThePower(number , power - 1)); else return 1; }
it's better if it is like this:
int byThePower(int number, int power)
{
if (power != 0)
return number * (byThePower(number , power - 1));
else
return 1;
}
can you explain how this code works: return number * (byThePower(number , power - 1));
return number * (byThePower(number , power - 1));I don't know how to explain it, so I would rather show how it process: Let say power = 4
Because of byThePower(number, 0) will probally return 1 so that.
Moreover, I prefer if (power >= 1) than if (power != 0) since it is safer. If you use if (power != 0) and the power is a negative number, then it leads to infinitive loop.
return number * (byThePower(number , power - 1));I don't know how to explain it, so I would rather show how it process: Let say power = 4
- Number to return = number * byThePower(number, 3)
- byThePower(number, 3) = number * byThePower(number, 2)
- byThePower(number, 2) = number * byThePower(number, 1)
- byThePower(number, 1) = number * byThePower(number, 0)
Because of
byThePower(number, 0)will probally return 1 so that.
- byThePower(number, 1) = number * 1 = number
- byThePower(number, 2) = number * byThePower(number, 1) = number * number
- byThePower(number, 3) = number * byThePower(number, 2) = number * number * number
- Number to return = number * byThePower(number, 3) = number * number * number * number
Moreover, I prefer
if (power >= 1)thanif (power != 0)since it is safer. If you useif (power != 0)and the power is a negative number, then it leads to infinitive loop.
Now i understand, thanks for helping me and giving information...
thanks for everything....
>>3^3 = 27.
that's the wrong answer, so you program will never get it (unless the program is wrong)
Wuh?
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