C Programming - Floating Point Issues - Discussion

Hi.

Rithvika normalization is one way to reduce the redundancy in table.

What do you mean by big endian and little endian ?

Hey can any one tell the procedure of normalisation?

Thanks preethi, visual, gaurav.

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

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

Thank preeti and pradeep.

How do normaliize ? please say anyone for me.

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

#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.

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.