Conversion of single precision double into a floating point

Question

How do I convert $0$ $00000001$ $00000000000000000000000$ into a floating point number?

Apparently the sign is +

I've started with $00000001$ which is 1 in decimal.

Then I applied formula $exponent = c -127 \Rightarrow exponent = -126$ and I got stuck.

What do you mean "convert into a floating point number"? What form should the result have, $1.234\times 2^p$, $1.234\times 10^e$? Or what? (Also: hidden bit) — Daniel Fischer, Sep 17 '13 at 09:24
$10^e$, but if you show me how to find the second one as well, I would be very grateful . — John Lennon, Sep 17 '13 at 10:04

score 1 · Answer 1 · answered Sep 17 '13 at 09:41

1

Hint: IEEE singles have a hidden implied highest bit in the mantissa. Therefore your number is the smallest normalised positive single (32-bit) number, and the answer is $2^{-126} \approx 0.117549435\cdot10^{-37}$

answered Sep 17 '13 at 09:41

gammatester

18,827

Warren Hill · Accepted Answer · 2013-09-17T09:55:32.100

1

I'm assuming this is IEEE754 single precision. In which case

The sign bit is 0 so the the number is positive

The exponent $e$ is $00000001_2 = 1$

The mantissa $m$ is $1.00000000000000000000000_2 = 1 $ The leading 1 is implied.

The number is thus

$m \times 2^{e-127} = 1 \times 2^{-126} \approx 1.17549435082228750796873653722 \times 10^{-38}$

There is an interactive tool here

And more information on Wikipedia here

edited Sep 17 '13 at 09:55

answered Sep 17 '13 at 09:50

Warren Hill

3,092

Small typo: The sign bit is 0. – gammatester Sep 17 '13 at 09:53
@gammatester Thanks I've edited my answer – Warren Hill Sep 17 '13 at 09:56

Conversion of single precision double into a floating point

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