Decimal to Binary

Im writing this converter and i dont see why the first time it gives me the right result
but after it gives me one 0 less every time.

EX. First time 8 --> 1000
After 8 --> 100

Code:

---

#include <iostream>
using namespace std;
#include <cstring>
#include <cstdlib>

int number, total, binary[8];
void prog();

int main()
{
      prog();
      return 0;
}

void prog()
{
  cout<<"Enter decimal to convert: ";
  cin>>number;
    while(number>0)
    {
        if((number%2)==0)
        {
            binary[total] = 0;
            number = number/2;
            total++; /* increasing by one each time will yield the
                        number of numbers in the array. */
        }
        else
        {
            binary[total] = 1;
            number = number/2;
            total++;
        }
    }
 
    total--;
   
    while(total>=0)
    {
        cout<<binary[total];
        total--;
    }

  cout<<endl<<endl;
  prog();
}

---

Instead of looping with recursion:

Code:

---

void foo()
{
    // do stuff
    foo(); 
}

---

loop with a real loop:

Code:

---

void foo()
{
    while(1)
    {
        // do stuff
    }
}

---

This way you don't run out of stack space.

Also, you need to initialize the total variable each time through the loop. This will likely fix your problem.

Thank you Bithub!

I changed both things you said. Initializing 'total' solved the problem.

I thought it was set back to zero each time in this loop. Apparently not.

Code:

---

if((number%2)==0)
        {
            binary[total] = 0;
            etc...

---

Here it assigns 0 to the first element of binary[].

your incrementing total and thats not set to 0 so its undefined value to start with

The first element of binary[total] is set to zero.
I tested it.

Code:

---

#include <iostream>
using namespace std;

int main()
{
    int binary[8], total=0;
    binary[total] = 3;

    for(int total = 0; total < 8; total++)
    cout << binary[total] <<endl;

return 0;
}

---

Umm, that program doesn't set the first element of binary[] to zero. Here's what it does:

• Allocate an array of 8 ints called binary, which are initialized to "random" values.
• Allocate a new int called total, which is initialized to zero.
• Set binary[total] to 3 -- i.e., set binary[0] or rather the first element in the array binary[] to 3. The remaining elements are uninitialized.
• Print out the elements of binary[]. You'll probably find that the first number is 3 and the other numbers seem pretty random.

Here was my output. (This is a system with 64-bit ints, so the numbers may have a larger range than your output.)

Code:

---

$ g++ init0.cpp -o init0
$ ./init0
3
5472000
0
-1079852504
134514489
10885717
6133204
-1079852520
$

---

Initialized to 0 or 3 doesnt make a difference. It was just an example.

My point was that its the first element that is initialized.

Code:

---

            number = number/2;
            total++; /* increasing by one each time will yield the
                        number of numbers in the array. */

---

This section of code is repeated twice - once in each side of the "else".

--
Mats

Quote:

---

Originally Posted by Ducky

Initialized to 0 or 3 doesnt make a difference. It was just an example.

My point was that its the first element that is initialized.

---

Considering that all variables being discussed are global, they should be all zero!

--
Mats

Quote:

---

Originally Posted by matsp

Code:

---

            number = number/2;
            total++; /* increasing by one each time will yield the
                        number of numbers in the array. */

---

This section of code is repeated twice - once in each side of the "else".

--
Mats

---

Which is normal considering that after division with 2 you get 1 or 0.

Quote:

---

Originally Posted by Ducky

Which is normal considering that after division with 2 you get 1 or 0.

---

Nah, you misunderstood the point of my post - you do the same thing on the if and else side of the if/else - that's called "code duplication", and should be avoided. You can just move those two lines of code to AFTER the if/else.

Of course, you can do ALL of this without if/else - but that's probably pushing it a bit far. [But if you want to, consider what the possible values of x%2 are?]

--
Mats

Basically you just shouldn't use globals for something as simple as that (you don't even want the variables to keep their value between calls!) and the program is broken up into functions in a rather poor way: main does too little, while prog does too much - part of it should be done in main. A better lay-out would be something like that (using std::string as a convenient type for storing the binary representation):

Code:

---

#include <iostream>
#include <string>

using namespace std;

//a function with a well-defined purpose
string to_binary(int num)
{
    //implementation of conversion
}

int main()
{
    while (true) { //or rather: while (cin)
        int number;
        cout<<"Enter decimal to convert: ";
        cin>>number;
        cout << to_binary(number) << endl << endl;
    }
}

---

Thanks for the help Mats and Anon.

I rewrote it, but i would still need to know how to put an int in a string.

Code:

---

#include <iostream>
#include <string>

using namespace std;

string to_binary(int number)
{
    string binary;

    while(number>=1)
    {
        number = number%2;
        binary = number;
    }
    return binary;
}

int main()
{
    while (cin)
    {
        int number;
        cout << "Enter decimal to convert: ";
        cin >> number;
        cout << to_binary(number) << endl << endl;

    }
}

---

You are probably going to need std::string a lot, so get familiar with it.

The function might look like this:

Code:

---

string to_binary(int number)
{
    string binary;

    while(number>=1)
    {
        //turn the numbers 0 or 1 into char's '0' or '1' and append to the string
        binary += (number % 2 + '0');
        number /= 2;
    }
   
    // the binary number is in reverse now
    //you could have added the characters to the beginning of the string earlier
    //but it's also simple just to reverse it:
    reverse(binary.begin(), binary.end()); //#include <algorithm>
    return binary;
}

---

For real programs you might also check out std::bitset.

Great, it works!

I dont know where they hide it but thats a cool trick to turn 'int' into 'string' with a '0'.

Thank you, Anon. You sure know how to recognize different types of trees from quite a long way away. :)

Quote:

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I dont know where they hide it but thats a cool trick to turn 'int' into 'string' with a '0'.

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It's rather well-known, and it turns digits into char's which you can append to a string.

You are complicating it too much.
Use the bitwise & operator, it seems to be indispensable in this program.

For example, to check if the number passed in parameters has the bit 0 as 1 or 0:

Code:

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<number_passed_in_params> & (int) pow(2, 0)

---

If the remainder of that operation is 0, than it has it as 0, if it is greater than 0, than 1.
Make a simple loop and through each iteration, apply that.

Except instead of pow one would use 1 << n for powers of two.