Using 64-bit IEEE 754 DOUBLE precision floating point with one(1) sign bit, eleven (11) exponent bits...

Using 64-bit IEEE 754 DOUBLE precision floating point with one(1) sign bit, eleven (11) exponent bits and fifty-two (52) mantissa bits, what is the decimal value of: 0xBFE4000000000000

Expert Solution

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 BFE4000000000000 to binary
B => 1011
F => 1111
E => 1110
4 => 0100
0 => 0000
0 => 0000
0 => 0000
0 => 0000
0 => 0000
0 => 0000
0 => 0000
0 => 0000
0 => 0000
0 => 0000
0 => 0000
0 => 0000
So, in binary BFE4000000000000 is 1011111111100100000000000000000000000000000000000000000000000000
1011111111100100000000000000000000000000000000000000000000000000
1 01111111110 0100000000000000000000000000000000000000000000000000
sign bit is 1(-ve)
exp bits are 01111111110
   => 01111111110
   => 0x2^10+1x2^9+1x2^8+1x2^7+1x2^6+1x2^5+1x2^4+1x2^3+1x2^2+1x2^1+0x2^0
   => 0x1024+1x512+1x256+1x128+1x64+1x32+1x16+1x8+1x4+1x2+0x1
   => 0+512+256+128+64+32+16+8+4+2+0
   => 1022
in decimal it is 1022
so, exponent/bias is 1022-1023 = -1
frac bits are 01

IEEE-754 Decimal value is 1.frac * 2^exponent
IEEE-754 Decimal value is 1.01 * 2^-1
1.01 in decimal is 1.25
   => 1.01
   => 1x2^0+0x2^-1+1x2^-2
   => 1x1+0x0.5+1x0.25
   => 1+0.0+0.25
   => 1.25
so, 1.25 * 2^-1 in decimal is 0.625
so, 0xBFE4000000000000 in IEEE-754 double precision format is -0.625
Answer: -0.625

venereology answered 8 months ago

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