binary to decimal

newbie0101 · 12-31-2011, 08:08 AM

hello guys, i am a beginner of C++ and to my mind came an idea to write a program that converts decimal to binary.
before i decided to write my own i was looking for some examples in google and found this:

Code:

#include <iostream>

using namespace std;

void binary(int);

int main() {
	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;
}

it works but i don't understand this line:

Code:

binary(number << 1)

i know that the binary function takes one argument to work with but what means this: "<< 1" ?

thanks for explanation

Doc CPU · 12-31-2011, 08:20 AM

Hi there,

Quote:

Originally Posted by newbie0101

but what means this: "<< 1" ?

a << 1 means: Take the value of a and shift-left by 1 bit position. Likewise, there's the >> operator, which is a right-shift.

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Proud · 12-31-2011, 08:22 AM

It looks like a recursive call to the binary() method, where the arguement is a bit-wise shifted value. A single shift will usually double or half a number in binary. Does this help you understand the algorithm?

lyle_s · 12-31-2011, 08:26 AM

Quote:

Originally Posted by newbie0101

it works but i don't understand this line:

Code:

binary(number << 1)

i know that the binary function takes one argument to work with but what means this: "<< 1" ?

In this context, "<<" is the left shift operator. It's shifting the bits that make up 'number' one to the left.

newbie0101 · 12-31-2011, 08:54 AM

thank you very much for such quick response but, why does it shift to the right by 1 bit and takes of the value of number actually ? (by the way there is a mistake in the code, there is number << 1 and correct is: number >> 1, was just experimentig)

lyle_s · 12-31-2011, 11:02 AM

Quote:

Originally Posted by newbie0101

why does it shift to the right by 1 bit and takes of the value of number actually ?

That line combines at least three concepts: recursion (the function called 'binary' calling itself), overloaded operators (>>), and temporary values (the fact that the inner call to 'binary' is called with an argument not stored in a named variable.)

I'll try to separate them out for you.

The line 'binary(number >> 1);' can be replaced with:

Code:

temp = number >> 1;
binary(temp);

If the line 'temp = number << 1;' still causes trouble, it works the same way as 'temp = number + 1;' in that plus operator takes two arguments and returns a result, just like right shift.

You've probably seen '<<' and '>>' used in input/output. The '<<' and '>>' operators are what is called "overloaded." This means that the functions that get called are dependant on the types the operator is called with.

So, in

Code:

cout << "happy times\n";

and

Code:

int shifted = 10 << 1;

completely different functions are being called, even though the same operator, <<, is used. You can write functions to overload operators to work with your own classes.

Hope this helps,
Lyle.

Doc CPU · 01-01-2012, 08:17 AM

Hi there,

Quote:

Originally Posted by lyle_s

That line combines at least three concepts: recursion (the function called 'binary' calling itself), overloaded operators (>>), and temporary values (the fact that the inner call to 'binary' is called with an argument not stored in a named variable.)

there is no overloading here, as the bit shift is the native operation of >> or <<.

Quote:

Originally Posted by lyle_s

You've probably seen '<<' and '>>' used in input/output. The '<<' and '>>' operators are what is called "overloaded."

Yes, they're overloaded for use with C++ iostreams as their first operand.

Quote:

Originally Posted by lyle_s

Code:

cout << "happy times\n";

and

Code:

int shifted = 10 << 1;

completely different functions are being called, even though the same operator, <<, is used.

Right, like the * operator has a different meaning depending on its right operand. If that is a pointer expression, the * is the dereferencing operator; if it's a number (or can be casted to one), it is the multiplication operator.
And that's not even C++, it's plain C.

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