Question

The 32-bit IEEE 754 single precision representation of a number is \(0xC2710000\). The number in decimal representation is ______. (rounded off to two decimal places)

Hint:

Split the 32 bits into 1 sign bit, 8 exponent bits, and 23 mantissa bits, then apply the IEEE 754 formula (-1)^S x 1.mantissa x 2^(exponent-127).

Answer 1

Correct Answer: -60.25

Step 1: Expand the hex value bit by bit.
$0xC2710000$ becomes, digit by digit, $1100\ 0010\ 0111\ 0001\ 0000\ 0000\ 0000\ 0000$ in binary, a full 32 bit word.

Step 2: Pull out the three fields.
The leftmost bit is the sign, $S=1$. The next 8 bits are the exponent field, $E=10000100_2$. The remaining 23 bits are the mantissa field, $M=11100010000000000000000$.

Step 3: Convert the exponent field to a number.
$$ E = 1\cdot2^7+0\cdot2^6+0\cdot2^5+0\cdot2^4+0\cdot2^3+1\cdot2^2+0\cdot2^1+0\cdot2^0 = 128+4=132 $$
Removing the bias of 127 for single precision gives the real exponent, $132-127=5$.

Step 4: Treat the mantissa as a fraction over $2^{23}$.
Only four bits of $M$ are set, at positions 1, 2, 3 and 7 after the point. As a fraction over $2^{23}$ this is
$$ M_{value} = \frac{2^{22}+2^{21}+2^{20}+2^{16}}{2^{23}} $$
Since $2^{22}+2^{21}+2^{20}+2^{16} = 4194304+2097152+1048576+65536=7405568$, and $2^{23}=8388608$,
$$ M_{value} = \frac{7405568}{8388608} = 0.8828125 $$

Step 5: Add the implied leading bit.
The significand is $1+M_{value}=1.8828125$.

Step 6: Scale by the exponent and apply the sign.
$$ (-1)^{1}\times 1.8828125\times 2^{5} = -1.8828125\times 32 $$

Step 7: Multiply out.
$1.8828125\times 32 = 60.25$, so with the negative sign the value is $-60.25$.
$$ \boxed{-60.25} $$