C Programming - Floating Point Issues - Discussion

@ Preethi: Thnx for providing nice answer...

clear to understand.

The above explanations are correct I would like to give an idea about another method of arrangement i.e 'Big-endian' here we can arrange 1234 as 3-4-1-2 or 2-1-4-3 in the memory addresses

Can any one please tel me how to do normalization ?

a=5.375 IS REPRESENTED AS '40 AC 00 00'. When stored in memory in case of little endian its stored as..

(eg.) 500 501 502 503
a = 00 00 AC 40

Now p=(char *)&a; means p will hold address of a and p is of char type so after typecasting it will hold 500 which hold 00, on incrementing it will print respective values.

#include<stdio.h>
#include<math.h>
int main()
{
float a=5.375;
char *p;
int i;
p = (char*)&a;
for(i=0; i<=3; i++)
printf("%02x\n", (unsigned char)p[i]);
return 0;
}

Please give me the description for how the above program is executing.

How will I understand the c program easily I feel the programs are very difficult. It is easy or not and please teach me.

Thank preeti and pradeep.

How do normaliize ? please say anyone for me.

Floating point base conversion of 5.375 would be
101.01100 00000000 00000000 00000000 00000000 00000000 00000000 10000000

(char*) will be make it as a character array.
5.375 = 101.01100 00000000 00000000 00000000 00000000 00000000 00000000 10000000

and bcoz of unsigned char, it wud be treated as,
p[0] = 00000000 00000000
p[1] = 00000000 00000000
p[2] = 10101100 00000000
p[2] = 00000000 10000000

=> in hexadecimal form (%02x) :
p[0] = 00
p[1] = 00
p[2] = ac
p[3] = 40

In little endian, lower order bytes will be stored in lower addresses. But how this is the answer?

Thanks preethi, visual, gaurav.