Hey guys! I'm trying to write a recursive function to take a number in binary form and change it to decimal. As of now I am having the dumbest problem. I can't seem to take 2 to a variable power. This is probably something so basic but I'm drawing a blank. (Oh, and I haven't written a clause to prevent any numbers outside of 0 &1 yet so please ignore that little part).

As of now, I can't even test if I have the right idea for my recursive function because I can't compile due to some issue with taking 2 to a particular power.

Here is my recursive function called binary:

int binary(const int data[], int position)
{
    int result;
    
    if(position==0)
    {
     result=data[position]*pow(2,position);
     }
     else
     {
     result=(data[position]*pow(2,position))+binary(data, position-1);
     }
     return result;
}

Let me know if you need my entire code! Thanks in advance for any suggestions. Oh, and I do have the cmath header in there so I know that much isn't the problem...:yawn:

And this is the error message I get:
D:\MyPrograms\Recursion>g++ Num4.cpp -o test4.exe
Num4.cpp: In function `int binary(const int*, int)':
Num4.cpp:44: error: call of overloaded `pow(int, int&)' is ambiguous
D:/mingw32_snapshot_02-10-2009/mingw/bin/../lib/gcc/mingw32/3.4.5/../../../../in
clude/math.h:156: note: candidates are: double pow(double, double)
D:/mingw32_snapshot_02-10-2009/mingw/bin/../lib/gcc/mingw32/3.4.5/../../../../in
clude/c++/3.4.5/cmath:361: note: long double std::pow(long doubl
...and it goes on...

Recommended Answers

All 10 Replies

This link shows the function prototype that pow function takes. Make
sure you match one of the function prototype.

So for example this code :

pow(2,2);

will not work since pow does not overload pow(int,int) . But this any of these will work :

pow(2.0,2.0); 
pow(2.0,2);
pow(2.0f,2.0f);
pow(2.0f,2);

hi firstPerson! I just saw your response but I already got it. Had to to do some typecasting.

The more pressing question now is why my function isn't giving me the correct answer when it's anything other than the base case.:-/
I "think" my logic is correct so I'm stumped...

Here it is now:

int binary(int data[], int position)
{
    double result;
    int decimal;
    
    if(position==0)
    {
     result=data[position]*pow((double)2,(double)position);
     }
     else
     {
     result=(data[position]*pow((double)2,(double)position))+binary(data, (double)position-1);
     }
     return result;
}

Okay, I've figured out what's going wrong! If I enter a number like 1, 111, 10101, etc, I get the correct answer. If I enter a number like 10(or anything unsymmetrical) I get the wrong answer. Soooo, my guess is that somehow I am going through the numbers in my array in the wrong order. *sighs*

If you have an 8 digit binary number, you want to have it so you're multiplying the value in position n with 2^(7-n). You could send in the size of the number in digits as a third parameter to your function that wouldn't change.

Hey jonsca! I see what you're saying but unfortunately the way I sorta "defined" position was merely length-1. I honestly have drawn out everything on paper and for the life of me cannot figure out why it basically computes the decimal number backwards. When I put in 100, I get the answer I would've gotten for 001 and vice versa. Crazy!

Here's my ENTIRE code:

#include <iostream>
#include <cmath>
using namespace std;

int binary(int data[], int position);

int main()
{
    int length;
    int tempNum;
    int i;
    
    cout<<"Enter the length of the binary number to be converted: "<<endl;
    cin>>length;
    
    int array[length];
    cout<<"Enter binary number one digit at a time. Hit return after each digit."<<endl;
    for(i=0; i<length; i++)
    {
     cin>>tempNum;
     array[i]=tempNum;
    }

    cout<<"The binary number you entered is: ";
    for(i=0; i<length; i++)
    {
     cout<<array[i];
    }
    cout<<endl;
    
    cout<<"The decimal form of the binary number you entered is: ";
    cout<<binary(array, length-1)<<endl;

    system("pause");
    return 0;
}
//*****************************************************************************
int binary(int data[], int position)
{
    double result;
    
    if(position==0)
    {
     result=data[position]*pow((double)2,(double)position);
     }
     else
     {
     result=(data[position]*pow((double)2,(double)position))+binary(data, (double)position-1);
     }
     return result;
}

Thanks again for the suggestions!

Hey back. If you trace your code you can see it: result=data[digits-1-position]*pow((double)2,(double)position); would do the trick (since you do want the low indexes to be multiplied by the higher values of 2^n and vice versa).

int binary(int data[], int position,int digits)
{
......
......
......       +binary(data,position-1,digits);  
}

digits gets passed all the way through and doesn't change with the recursion. You always have it around. There might be a better way but that was my thought.

Ah, seems like it should work but I actually had tried that too. It gets rid of the problem with "unsymmetrical" binary numbers but anything with leading or trailing zeroes is always off by 1(either 1 too much or 1 too little).
This is what I did:

#include <iostream>
#include <cmath>
using namespace std;

int binary(int data[], int position, int length);

int main()
{
    int length;
    int tempNum;
    int i;
    
    cout<<"Enter the length of the binary number to be converted: "<<endl;
    cin>>length;
    
    int array[length];
    cout<<"Enter binary number one digit at a time. Hit return after each digit."<<endl;
    for(i=0; i<length; i++)
    {
     cin>>tempNum;
     array[i]=tempNum;
    }

    cout<<"The binary number you entered is: ";
    for(i=0; i<length; i++)
    {
     cout<<array[i];
    }
    cout<<endl;
    cout<<endl;
    
    cout<<"The decimal form of the binary number you entered is: ";
    cout<<binary(array, length-1, length)<<endl;

    system("pause");
    return 0;
}
//*****************************************************************************
int binary(int data[], int position, int length)
{
    double result;
    
    if(position==0)
    {
     result=data[position]*pow((double)2,(double)position);
     }
     else
     {
     result=(data[length-1-position]*pow((double)2,(double)position))+binary(data, position-1, length);
     }
     return result;
}

For example: 10, 01, 1000, 0001, etc...These numbers are always off.

Correction/clarification: if there are leading zeroes, the final answer is always 1 less than what it should be. If there are trailing zeroes, the final answer is always 1 more than what it should be.

You didn't change your base case to reflect your new changes.

Be a part of the DaniWeb community

We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.