ieee representation of floating point numbers

IEEE REPRESENTATION OF FLOATING POINT NUMBER

An IEEE-754 float (4 bytes) or double (8 bytes) has three components (there is also an analogous 96-bit extended-precision format under IEEE-854):

-> a sign bit telling whether the number is positive or negative, 

->an exponent giving its order of magnitude,

->an mantissa specifying the actual digits of the number. 

Using single-precision floats(32 bit) as an example, here is the bit layout:-

we have to represent it in form of ->1.M *2^(+/-E)

where:-

 ->M is the mantissa which is to be represented in 23 bit for single precision number.

->E is the exponent  which is to be represented in 8 bit for single precision number.

eg 1:

now suppose the floating point no: is 11.27

->signbit in 1 bit is: 1

->mantissa in 23 bit is: :0 1 1 0 1 0 0 0 1 0 1 0 0 0 1 1 1 1 0 1 0 1 1 

->exponent is 8 bit: 10000010

eg 2:

now suppose the floating point no: is 157.38
->signbit in 1 bit is: 1

->mantissa in 23 bit is: :0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 0 1 0 0 0 1 1 1 

->exponent is 8 bit: 10000110

C CODE FOR IEEE REPRESENTATION OF FLOATING POINT NUMBER

#include<stdio.h>

#include<stdlib.h>

#include<math.h>

static int r[23],mant[24];

int dec2bin(int n,int *arr)

{

int i,p;

p=n;

for(i=0;i<23;i++)

arr[i]=0;

i=22;

do

{

arr[i]=n%2;

n=n/2;

i--;

}

while(n!=0);

if(p<0)

{

printf("****negative number****\n");

printf("do enter a positive number\n");

}

else

{

printf("the binary equivalent of %d is:",p);

for(i=0;i<23;i++)

printf("%d",arr[i]);

}

printf("\n");

}

int dec2bin4exp(int n,int *arr)

{

        int i,p;

        p=n;

        for(i=0;i<8;i++)

        arr[i]=0;

        i=7;

        do

        {

                arr[i]=n%2;

                n=n/2;

                i--;

        }

        while(n!=0);

                printf("the binary equivalent of %d is ",p);

printf("(i.e the exponent part in 8 bit is:)");

                for(i=0;i<8;i++)

                printf("%d",arr[i]);

        

                printf("\n");

}

int sign(float n)

{

int sign;

if(n>0)

sign=0;

else

sign=1;

printf("the signbit=%d\n",sign);

}

int rep(int *arr)

{

    

     int i,j,z=0,t=0,l,flag=0,power,sign,bias,k=0;

     for(i=0;i<23;i++)

     {

            if(arr[i]==1)

            z=i;

   if(z>0)

   {

for(l=z+1;l<23;l++)

 mant[k++]=arr[l];

     break;

   }

     }

      bias=k+127;

      //printf("K=%d",k);

for(j=0;j<23;j++)

{

mant[k++]=r[j];

    }

printf("the mantissa part in 23 bit is:");

for(i=0;i<23;i++)

{

printf("%d ",mant[i]);

}

printf("\n");

printf("------------------------------------------------------\n");

printf(" bias value is=%d\n",bias);

dec2bin4exp(bias,r);

}

void pt2bin(double d)

{

int i;

static double ar[23],f=0.0;

ar[0]=d;

// printf("\nar=%f\n",ar[0]);

for(i=0;i<23;i++)

{

f=ar[i] *2;

r[i]=(int)f;

ar[i+1]=f-r[i];

}

printf("The binary equivalent of %lf is:",d);

int ii;

for(ii=0;ii<23;ii++)

printf("%d ",r[ii]);

printf("\n");

}

int main()

{

double f,deci;

int real,arr[23],arr1[5];

printf("enter a floating point number\n");

scanf("%lf",&f);

real=f;

deci=f-real;

printf("------------------------------------------------------\n");

printf("the real part of the number is:=%d\n",real);

printf("the decimal part of the number is:=%lf\n",deci);

printf("------------------------------------------------------\n");

dec2bin(real,arr);

pt2bin(deci);

printf("------------------------------------------------------\n");

sign(f);

rep(arr);

printf("------------------------------------------------------\n");

return 0;

}