Find the largest power of 2 which is smaller than your number, e.g if you start with x = 10.0 then 23 = 8, so the exponent is 3. The exponent is biased by 127 so this means the exponent will be represented as 127 + 3 = 130. The mantissa is then 10.0/8 = 1.25. The 1 is implicit so we just need to represent 0.25, which is 010 0000 0000 0000 0000 0000 when expressed as a 23 bit unsigned fractional quantity. The sign bit is 0 for positive. So we have:
s | exp [130] | mantissa [(1).25] |
0 | 100 0001 0 | 010 0000 0000 0000 0000 0000 |
0x41200000
You can test the representation with a simple C program, e.g.
#include <stdio.h>
typedef union
{
int i;
float f;
} U;
int main(void)
{
U u;
u.f = 10.0;
printf("%g = %#x\n", u.f, u.i);
return 0;
}
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Find the largest power of 2 which is smaller than your number, e.g if you start with x = 10.0 then 23 = 8, so the exponent is 3. The exponent is biased by 127 so this means the exponent will be represented as 127 + 3 = 130. The mantissa is then 10.0/8 = 1.25. The 1 is implicit so we just need to represent 0.25, which is 010 0000 0000 0000 0000 0000 when expressed as a 23 bit unsigned fractional quantity. The sign bit is 0 for positive. So we have:
s | exp [130] | mantissa [(1).25] |
0 | 100 0001 0 | 010 0000 0000 0000 0000 0000 |
0x41200000
You can test the representation with a simple C program, e.g.
#include <stdio.h>
typedef union
{
int i;
float f;
} U;
int main(void)
{
U u;
u.f = 10.0;
printf("%g = %#x\n", u.f, u.i);
return 0;
}
Take a number 172.625.This number is Base10 format.
Convert this format is in base2 format For this, first convert 172 in to binary format
128 64 32 16 8 4 2 1
1 0 1 0 1 1 0 0
172=10101100
Convert 0.625 in to binary format
0.625*2=1.250 1
0.250*2=.50 0
0.50*2=1.0 1
0.625=101
Binary format of 172.625=10101100.101. This is in base2 format 10101100*2
Shifting this binary number
1.0101100*2 **7 Normalized
1.0101100 is mantissa
2 **7 is exponent
add exponent 127 7+127=134
convert 134 in to binary format
134=10000110
The number is positive so sign of the number 0
0 |10000110 |01011001010000000000000
Explanation: The high order of bit is the sign of the number. number is stored in a sign magnitude format. The exponent is stored in 8 bit field format biased by 127 to the exponent The digit to the right of the binary point stored in the low order of 23 bit. NOTE---This format is IEEE 32 bit floating point format
10/32 = 5/16 = 5•2−4 = 1.25•2−2 = 1.012•2−2.
The sign is +, the exponent is −2, and the significand is 1.012.
A positive sign is encoded as 0.
Exponent −2 is encoded as −2 + 127 = 125 = 011111012.
Significand 1.012 is 1.010000000000000000000002, and it is encoded using the last 23 bits, 010000000000000000000002.
Putting these together, the IEEE-754 encoding is 0 01111101 01000000000000000000000. To convert to hexadecimal, first organize into groups of four bits: 0011 1110 1010 0000 0000 0000 0000 0000. Then the hexadecimal can be easily read: 3EA0000016.
I see it like this:
10/32 = // input
10/2^5 = // convert division by power of 2 to bitshift
1010b >> 5 =
.01010b // fractional result
--^-------------------------------------------------------------
|
first nonzero bit is the exponent position and start of mantissa
----------------------------------------------------------------
man = (1)010b // first one is implicit
exp = -2 + 127 = 125 // position from decimal point + bias
sign = 0 // non negative
----------------------------------------------------------------
0 01111101 01000000000000000000000 b
^ ^ ^
| | mantissa + zero padding
| exp
sign
----------------------------------------------------------------
0011 1110 1010 0000 0000 0000 0000 0000 b
3 E A 0 0 0 0 0 h
----------------------------------------------------------------
3EA00000h
Yes the answer of Eric Postpischil is the same approach (+1 btw) but I didn't like the formating as it was not clear from a first look what to do without proper reading the text.