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 10 of 12.
Preeti said:
1 decade ago
Thank you preeti:).
Chaitanya@rgukt said:
1 decade ago
Can any please explain How to do normalization of floating point?
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
Suma said:
1 decade ago
@Shilpa m raj.
1.01011*2^2 = 1.01011*100 where 2 is true exponent.
I didn't understand, how *2^2 become *100.
1.01011*2^2 = 1.01011*100 where 2 is true exponent.
I didn't understand, how *2^2 become *100.
Praful said:
1 decade ago
I think this value stored in this manner.
0100 0000 1010 1100 0000 0000 0000 0000 which is 40ac0000.
float taken 4 byte in memory in program p=(char*)&a casting this to char type nw p(it char pointer its point to i byte value) point to lower byte of the a(5.375) which is 0000 0000.
p[0]-> contain the fist lower byte data 00(because print bye the hexa format specifier withe 2 width)
p[1]-> contain 00
p[2]-> contain ac
p[3]-> contain 40
Then output become ...0000ac40.
0100 0000 1010 1100 0000 0000 0000 0000 which is 40ac0000.
float taken 4 byte in memory in program p=(char*)&a casting this to char type nw p(it char pointer its point to i byte value) point to lower byte of the a(5.375) which is 0000 0000.
p[0]-> contain the fist lower byte data 00(because print bye the hexa format specifier withe 2 width)
p[1]-> contain 00
p[2]-> contain ac
p[3]-> contain 40
Then output become ...0000ac40.
Piya said:
1 decade ago
Little Endian means that the lower order byte of the number is stored in memory at the lowest address, and the higher order byte is stored at the highest address. That is, the little end comes first.
For example, a 4 byte, 32-bit integer.
Byte3 Byte2 Byte1 Byte0.
will be arranged in memory as follows:
Base_Address+0 Byte0.
Base_Address+1 Byte1.
Base_Address+2 Byte2.
Base_Address+3 Byte3.
Example :Intel processors use "Little Endian" byte order.
"Big Endian" means that the higher order byte of the number is stored in memory at the lowest address, and the lower order byte at the highest address. The big end comes first.
Base_Address+0 Byte3.
Base_Address+1 Byte2.
Base_Address+2 Byte1.
Base_Address+3 Byte0.
Motorola, Solaris processors use "Big Endian" byte order.
For example, a 4 byte, 32-bit integer.
Byte3 Byte2 Byte1 Byte0.
will be arranged in memory as follows:
Base_Address+0 Byte0.
Base_Address+1 Byte1.
Base_Address+2 Byte2.
Base_Address+3 Byte3.
Example :Intel processors use "Little Endian" byte order.
"Big Endian" means that the higher order byte of the number is stored in memory at the lowest address, and the lower order byte at the highest address. The big end comes first.
Base_Address+0 Byte3.
Base_Address+1 Byte2.
Base_Address+2 Byte1.
Base_Address+3 Byte0.
Motorola, Solaris processors use "Big Endian" byte order.
Sudhir said:
1 decade ago
Here what is mean by printf("%02x\n", (unsigned char)p[i]); statement, please explain me.
Sayan bose said:
1 decade ago
How is hexadecimal operation taking place in a compiler?
Suresh Gupta said:
1 decade ago
We are use database in C or not ?
Sameer Sarkaar said:
1 decade ago
What is little endian?
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