Question

What decimal number would the IEEE 754 single precision floating point number 0xC40E5B00 (this is in hex) be? Write your final answer in scientific notation as m x 10 ^p where p is an integer.

Answer 1

Hexadecimal     Binary
    0           0000
    1           0001
    2           0010
    3           0011
    4           0100
    5           0101
    6           0110
    7           0111
    8           1000
    9           1001
    A           1010
    B           1011
    C           1100
    D           1101
    E           1110
    F           1111
Use this table to convert from hexadecimal to binary
Converting C40E5B00 to binary
C => 1100
4 => 0100
0 => 0000
E => 1110
5 => 0101
B => 1011
0 => 0000
0 => 0000
So, in binary C40E5B00 is 11000100000011100101101100000000
11000100000011100101101100000000
1 10001000 00011100101101100000000
sign bit is 1(-ve)
exp bits are 10001000
   => 10001000
   => 1x2^7+0x2^6+0x2^5+0x2^4+1x2^3+0x2^2+0x2^1+0x2^0
   => 1x128+0x64+0x32+0x16+1x8+0x4+0x2+0x1
   => 128+0+0+0+8+0+0+0
   => 136
in decimal it is 136
so, exponent/bias is 136-127 = 9
frac bits are 000111001011011

IEEE-754 Decimal value is 1.frac * 2^exponent
IEEE-754 Decimal value is 1.000111001011011 * 2^9
1.000111001011011 in decimal is 1.112152099609375
   => 1.000111001011011
   => 1x2^0+0x2^-1+0x2^-2+0x2^-3+1x2^-4+1x2^-5+1x2^-6+0x2^-7+0x2^-8+1x2^-9+0x2^-10+1x2^-11+1x2^-12+0x2^-13+1x2^-14+1x2^-15
   => 1x1+0x0.5+0x0.25+0x0.125+1x0.0625+1x0.03125+1x0.015625+0x0.0078125+0x0.00390625+1x0.001953125+0x0.0009765625+1x0.00048828125+1x0.000244140625+0x0.0001220703125+1x6.103515625e-05+1x3.0517578125e-05
   => 1+0.0+0.0+0.0+0.0625+0.03125+0.015625+0.0+0.0+0.001953125+0.0+0.00048828125+0.000244140625+0.0+6.103515625e-05+3.0517578125e-05
   => 1.112152099609375
so, 1.112152099609375 * 2^9 in decimal is 569.421875
so, 11000100000011100101101100000000 in IEEE-754 single precision format is -569.421875
Answer: -5.69421875 * 10^2