C Programming - Floating Point Issues - Discussion
Discussion Forum : Floating Point Issues - General Questions (Q.No. 3)
3.
If the binary eauivalent of 5.375 in normalised form is 0100 0000 1010 1100 0000 0000 0000 0000, what will be the output of the program (on intel machine)?
#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;
}
Discussion:
117 comments Page 4 of 12.
Shilpa M Raj said:
1 decade ago
Normalize:
Given Floating point Number = 5.375
In binary:
5 = (101) and
0.375 = (011)
{ 0.375*2 = 0.750 --> 0
0.750*2 = 1.500 --> 1
0.500*2 = 1.000 --> 1 }
i.e., 5.375 = 101.011
For single precision representation, decimal point is shifted by 2 times.
i.e., 1.01011*2^2 = 1.01011*100 where 2 is true exponent.
According to IEEE, the representation has three fields:
----------------------------
| S | E | F |
----------------------------
S is one bit representing the sign of the number (0/1 & hidden bit)
E is an 8-bit biased integer representing the exponent
F is an unsigned integer
For single precision representation (the emphasis in this class)
n = 23
bias = 127
2 is the true exponent. For the standard form, it needs to be in 8-bit, biased-127 representation.
2
+127
-----
129
129 in 8-bit, unsigned representation is 1000 0001
The mantissa stored (F) is the stuff to the right of the radix point in the normalized form. We need 23 bits of it.
010110 00000000000000000
Put it all together (and include the correct sign bit):
S E F
0 1000 0001 010110 00000000000000000
The values are often given in hex, so here it is
0100 0000 1010 1100 0000 0000 0000 0000
0x 4 0 A C 0 0 0 0
Given Floating point Number = 5.375
In binary:
5 = (101) and
0.375 = (011)
{ 0.375*2 = 0.750 --> 0
0.750*2 = 1.500 --> 1
0.500*2 = 1.000 --> 1 }
i.e., 5.375 = 101.011
For single precision representation, decimal point is shifted by 2 times.
i.e., 1.01011*2^2 = 1.01011*100 where 2 is true exponent.
According to IEEE, the representation has three fields:
----------------------------
| S | E | F |
----------------------------
S is one bit representing the sign of the number (0/1 & hidden bit)
E is an 8-bit biased integer representing the exponent
F is an unsigned integer
For single precision representation (the emphasis in this class)
n = 23
bias = 127
2 is the true exponent. For the standard form, it needs to be in 8-bit, biased-127 representation.
2
+127
-----
129
129 in 8-bit, unsigned representation is 1000 0001
The mantissa stored (F) is the stuff to the right of the radix point in the normalized form. We need 23 bits of it.
010110 00000000000000000
Put it all together (and include the correct sign bit):
S E F
0 1000 0001 010110 00000000000000000
The values are often given in hex, so here it is
0100 0000 1010 1100 0000 0000 0000 0000
0x 4 0 A C 0 0 0 0
Chaitanya@rgukt said:
1 decade ago
Can any please explain How to do normalization of floating point?
Preeti said:
1 decade ago
Thank you preeti:).
Shyam said:
1 decade ago
Understand the memory you will understand it easy
Used is little endian type of memory storage i.e., higher byte at lower address
Data:40 AC 00 00
100:40
101:AC
102:00
103:00
Now pointer to memory location points to 103
If it is used as character pointer it grabs a byte as char pointer is capable of handling only value of its size
On the other end if a integer pointer is used it would give output same as it is
For detailed view try this example,
int x=Ox12345678;
char *cp;
short int *sip;
int *ip;
ip=sip=cp=&x;
printf("%x %x %x",*cp,*sip,*ip);
O/P:0x78 0x5678 0x12345678
Used is little endian type of memory storage i.e., higher byte at lower address
Data:40 AC 00 00
100:40
101:AC
102:00
103:00
Now pointer to memory location points to 103
If it is used as character pointer it grabs a byte as char pointer is capable of handling only value of its size
On the other end if a integer pointer is used it would give output same as it is
For detailed view try this example,
int x=Ox12345678;
char *cp;
short int *sip;
int *ip;
ip=sip=cp=&x;
printf("%x %x %x",*cp,*sip,*ip);
O/P:0x78 0x5678 0x12345678
Nazia said:
1 decade ago
The format string used for each of the prints is %02X, which I've always interpreted as 'print the supplied int as a hexadecimal value with at least two digits'.
Suvidha said:
1 decade ago
In the statement p= (char*) &a; we are doing type casting since p is a char pointer you cannot assign float variable to it a pointer variable can hold the address of another variable only if the variable has the same data type of that as pointer variable.
Bhaskar said:
1 decade ago
Binary equivalent of 5.375 in normalised form is
0100 -> 4
0000 -> 0
1010 -> A
1100 -> C
0000 -> 0
0000 -> 0
0000 -> 0
0000 -> 0
Since the PC's (intel processors) use " LITTLE ENDIAN" byte order, the higher order byte of number is stored in lowest address. The result is written from bottom to top.
0100 -> 4
0000 -> 0
1010 -> A
1100 -> C
0000 -> 0
0000 -> 0
0000 -> 0
0000 -> 0
Since the PC's (intel processors) use " LITTLE ENDIAN" byte order, the higher order byte of number is stored in lowest address. The result is written from bottom to top.
Ragini said:
1 decade ago
Thanks preeti.
Er.Nitesh mathur said:
1 decade ago
Preethi is wrong because. Defination is given by preethi is totally wrong.
" Big ENDIAN" byte order-> the higher order byte of number is stored in lowest address. The result is written from top to bottom.
" LITTLE ENDIAN"-> byte order, the higher order byte of number is stored in higher address. The result is written from bottom to top.
" Big ENDIAN" byte order-> the higher order byte of number is stored in lowest address. The result is written from top to bottom.
" LITTLE ENDIAN"-> byte order, the higher order byte of number is stored in higher address. The result is written from bottom to top.
Chithra said:
1 decade ago
Thanks preethi and pradeep.
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