Conversion from decimal to bianary

This a program for conversion from decimal to binary where I have some problem to understand the binary function---

Code:

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#include <iostream>

using namespace std;

void binary(int);

int main(void) {
        int number;

        cout << "Please enter a positive integer: ";
        cin >> number;
        if (number < 0)
                cout << "That is not a positive integer.\n";
        else {
                cout << number << " converted to binary is: ";
                binary(number);
                cout << endl;
        }
}

void binary(int number) {
        int remainder;

        if(number <= 1) {
                cout << number;
                return;
        }


        remainder = number%2;
        binary(number >> 1);
        cout << remainder;
}

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Say, I would like to convert Decimal 11 to binary number. The program is running properly and giving the answer 1011. I guess that the binary function will be started with 11 and goes to

remainder = number%2;

The reminder will be 1 but the number is still 11 , right ? I did not understand the function here : binary(number >> 1) due to the ">>" sign.

Anyone explain it for me a little ?

Well this function uses recursion to repeat the code until it has displayed all of the bits. The call binary(number >> 1); is where the recursion takes place, and binary's argument provides the new number to operate on and display. The code number >> 1 is a bitwise operation. If you can imagine a series of bits, each shift by one removes the rightmost bit.

Excuse me, but I don't know what is bitwise operation. binary(number >> 1) here the number is 11, right ? Then ,is it goes top of the program --> void binary(int number) ?

The diagram on this page might help.

Thanks bernt :) I will have a look and write again.

SourceForge.net: Do not remove parameter names - cpwiki

Sorry for late reply. I need to do something yesterday so was not able to write back. I got the term ---> (number >> 1)

Shifting left by n bits on a signed or unsigned binary number has the effect of multiplying it by 2n. Shifting right by n bits on an unsigned binary number has the effect of dividing it by 2n (rounding towards 0)

For the number 11, its goes to the function first with 5 then, 2 and 1. My question is then it should print the binary as 1101, right ? Also, after the term remainder = number%2, every-time due to recursion its again go in the top to use the function. When the
cout << remainder;

It would help me if someone have time to explain me this. Thanks and wishes from Munich :)

Another question, what is the data type of binary numbers ?

Say, the decimal number 167 has data type int. Whats for 0 or 1 or something more say, 1100101 ?

There is no type for binary numbers since memory is byte-addressed. Hence, the smallest unit the processor works with is a byte (ie char).
You can only treat existing data types as binary by using bit operations.

Well, perhaps I should expand upon that. All data types are in some way or form binary since all data processed by a computer is a number. A character is merely a number that is interpreted differently when printed to screen.
Hence, all data has a binary representation, but there is no data type where you can directly store 0s and 1s in C++.

Thanks Elysia :) I had another question that is more importation for me to understand the program. Could you please explain me the binary() function to me properly ?

Nope, it's unintuitive. Taking the reminder may work in math, but in computer science, there is a way better and intuitive way.

EDIT: Here's how you do it with computers:

Code:

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#include <iostream>

int main()
{
        unsigned int num = 1234567890;
        int shift = 32;

        while (shift-- > 0)
                std::cout << ((num >> shift) & 0x1);
}

---

Sorry but I did not understand this code. It seems that if I put a decimal as number then it returns the binary with zeros before. In total 32 digits are printed. However, thats the meaning of shift-- > 0. Also, I did not understand the form: std::cout << ((num >> shift) & 0x1); specially why do you need to put std infornt.

I would rather prefer the explanation of the function binary () in the code posted before.

It assumes a number is 4 bytes, which means it has 32 bits. Therefore, it prints all those 32 bits. You can get rid of the leading 0s if you want.
std:: is part of namespaces. Read up about them.

Something about the code I posted before ? :)

Quote:

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For the number 11, its goes to the function first with 5 then, 2 and 1. My question is then it should print the binary as 1101, right ? Also, after the term remainder = number%2, every-time due to recursion its again go in the top to use the function. When the
cout << remainder;

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Quote:

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Originally Posted by chaklader

For the number 11, its goes to the function first with 5 then, 2 and 1. My question is then it should print the binary as 1101, right ? Also, after the term remainder = number%2, every-time due to recursion its again go in the top to use the function. When the
cout << remainder;

It would help me if someone have time to explain me this. Thanks and wishes from Munich :)

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Let's discuss the recursion: like you noted, the number 11 would be divided, first resulting in 5 then 2, then 1. The computer will follow the calls of binary in order.

binary( 11 );
binary( 5 );
binary( 2 );
binary( 1 );

When the calls return, the stack is unwound in reverse order, but in this case it doesn't affect the outcome.

The remainder is a way of finding the highest bit for the current number, and by shifting the number left, you work with the next highest bit and so forth.