CODEWITHSUNDEEP
floating-pointbinary
user3247608
user3247608

Reputation: 603

How to convert 23.5 into binary using 1 byte for the mantissa and 1 byte for the exponent

I tried and this is what i've got so far:

10111.100 = 1.0111100 x 2 ^ 100

Giving 10111100 for the mantissa and 00000100 for the exponent, but i think i may have got it wrong.

Upvotes: 0

Views: 963

Answers (1)

Rudy Velthuis
Rudy Velthuis

Reputation: 28826

23.5 is the same as 47 x 2^-1. 47 is binary 00101111. Now simply shift that right (and increment the exponent accordingly) until you get 1.0111100 and an exponent of 4 (binary 00000100). I assume you have some kind of bias for the exponent (probably 128) to cater for negative exponents, so add that.

In other words

23.5 = 47 * 2^-1 --> 00101111. exp -1
                     0010111.1 exp 0
                     001011.11 exp 1
                     00101.111 exp 2
                     0010.1111 exp 3
                     001.01111 exp 4 --> m=00101111 e=00000100

But normally, such value are normalized to set the top bit to 1:

1.01111000 exp 4.

And the exponent gets a bias, so negative values are turned into positive ones, say 128. Add 4 and you get:

e=10000100

The top mantissa (a.k.a. significand) bit is always 1, so it can be discarded, so now you have:

m=01111000 e=10000100

This is how I would store such a 16 bit floating binary point value.

Upvotes: 1

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