My previous post was based on converting a decimal number into its binary equivalent. Contrary to that, the program given below

Now for those who are not familiar with the concept of binary numbers, please read the following post : decimal to binary conversion.

Now, the method works like this. Just traverse the binary number step by step, starting from 0 to n-1, where n is the most significant bit(MSB) ,

1 0 0 0 ==> going from right to left

0 x 2^0 = 0

0 x 2^1 = 0;

0 x 2^2 = 0;

1 x 2^3 = 8;

now add them together i.e 0+0+0+8 = 8; this the decimal equivalent of 1000. Please read the program below to have a better understanding how the concept**converts binary numbers into their decimal equivalents**. That is, it**converts binary numbers such as 1000 into its corresponding decimal equivalent i.e 8.**Now for those who are not familiar with the concept of binary numbers, please read the following post : decimal to binary conversion.

Now, the method works like this. Just traverse the binary number step by step, starting from 0 to n-1, where n is the most significant bit(MSB) ,

**multiply them with 2 with raising powers and add the sum together**. E.g to convert 1000(which is binary equivalent of 8), just do the following1 0 0 0 ==> going from right to left

0 x 2^0 = 0

0 x 2^1 = 0;

0 x 2^2 = 0;

1 x 2^3 = 8;

work.

**Note : The program works only for 16-bit binary numbers(non-floating) or less. Leave a comment if anything is not clear. You are bound to receive a reply.**

**// Program to convert binary to its decimal equivalent**

#include <iostream>

#include <math.h>

int main()

{

int x;

int i=0,sum = 0;

// prompts the user to input a 16-bit binary number

std::cout<<" Enter the binary number (16-bit) : ";

std::cin>>x;

while ( i != 16 ) // runs 16 times

{

sum += (x%10) * pow(2,i);

x = x/10;

i++;

}

std::cout<<"\n The decimal equivalent is : "<<sum;

return 0;

}

**------ OUTPUT ------**

**------ Programming Advice and Tips ------**

**1 : Note : The above program only accepts 16-bit binary numbers as inputs. This limit can be increased by increasing the number of loop executions by**

changing the value in the while loop from 16 to say 32. (be sure it is a valid number which is a factor of 2 ).

2 : This program is the most basic method to do this type of conversion. However, there are a number of algorithms to do the same, so just GOOGLE out some

of them. Try it out, there are a variety of programs that do the same conversion in different ways.

changing the value in the while loop from 16 to say 32. (be sure it is a valid number which is a factor of 2 ).

2 : This program is the most basic method to do this type of conversion. However, there are a number of algorithms to do the same, so just GOOGLE out some

of them. Try it out, there are a variety of programs that do the same conversion in different ways.

**Please do comment if you don't understand any part or want to know more or just want to say thanks. I love programming and love to teach my friends. Your suggestions and appreciation will make this blog much better.**

**------ Related Posts ------**

**1 : Decimal to binary conversion (v1.0) : http://programsplusplus.blogspot.in/2012/04/my-apologies-to-my-readers-for-posting.html**

**2 : Decimal to binart conversion (v2.0) : http://programsplusplus.blogspot.in/2012/04/decimal-to-binary-conversion-program.html**

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ReplyDeletecheck this blog

I didn't understand why there is a"+" next to the variable "sum" -->sum +=

ReplyDeletei don't understand the meaning of sum+=(x%10)*pow(2,i) plz elaborate it

ReplyDeleteThank you ! This was the clean and simple answer I was looking for.

ReplyDeletethe program is incorrect as when you enter binary number as 0010 it gives output as 8

ReplyDeleteplease provide with an efficient program

ReplyDelete