Binary to Decimal Conversion in C Programming

 

// BINARY to DECIMAL

#include <stdio.h>

#include <math.h> 

#include <string.h> 

int main()

{

int n=0,POWER=0,i,j;

char ar1[20];

char ar2[2] = {'0','1'}; //ar2[0]=0, ar2[1]=1

printf("Enter any Binary Number : ");

scanf("%s",&ar1);

for(i=strlen(ar1)-1;i>=0;i--)

//let entered number = 1011, then i=4-1=3;i>=0;i--

//it means i=3,2,1,0 

{

for(j=0;j<2;j++)

//j=0,1

{

if(ar1[i]==ar2[j])

{

n=n+j*pow(2,POWER);

}

}

POWER++;

/*

in first step

ar1[3]=1,ar2[0]=0 and ar2[1]=1

therefore ar1[3]=ar2[1]

it means n = 0+1*pow(2,POWER)= 0+1*pow(2,0) = 0+1*1 = 1

in second step

ar1[2]=1,ar2[0]=0 and ar2[1]=1

therefore ar1[2]=ar2[1]

it means n = 1+1*pow(2,POWER)= 1+1*pow(2,1) = 1+1*2 = 3

in third step

ar1[1]=0,ar2[0]=0 and ar2[1]=1

therefore ar1[1]=ar2[0]

it means n = 3+0*pow(2,POWER)= 3+0*pow(2,2) = 3+0*4 = 3

in fourth step

ar1[0]=1,ar2[0]=0 and ar2[1]=1

therefore ar1[0]=ar2[1]

it means n = 3+1*pow(2,POWER)= 3+1*pow(2,3) = 3+1*8 = 11

now n = 11

*/

}

printf("Decimal Number = %d",n);

}

//Another Method

//Binary to Decimal

#include <stdio.h>  

int main()

{

int n,dec=0,temp=1,REM;

printf("Enter any Binary Number : ");

scanf("%d",&n);

while(n>0)

{

REM = n%10;

dec = dec + REM*temp;

n = n/10;

temp = temp*2;

}

/*

let n = 11101

step 1:

11101>0 true

REM = 11101%10 = 1

dec = 0 + 1*1 = 0 + 1 = 1

n = 11101/10 = 1110

temp = 1*2 = 2

step 2:

1110>0 true

REM = 1110%10 = 0

dec = 1 + 0*2 = 1 + 0 = 1

n = 1110/10 = 111

temp = 2*2 = 4

step 3:

111>0 true

REM = 111%10 = 1

dec = 1 + 1*4 = 1 + 4 = 5

n = 111/10 = 11

temp = 4*2 = 8

step 4:

11>0 true

REM = 11%10 = 1

dec = 5 + 1*8 = 5 + 8 = 13

n = 11/10 = 1

temp = 8*2 = 16

step 4:

1>0 true

REM = 1%10 = 1

dec = 13 + 1*16 = 13 + 16 = 29

n = 1/10 = 0

temp = 16*2 = 32

step 5:

0>0 false

*/

printf("Decimal Number : %d",dec);

return 0;

}

**Author**

Sukhvir Singh

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