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Hi I have some trouble with how should I add two 8-bit floating points with different signs. The question is here,

1 100 1100 + 0 101 1011 =

Thee 1st bit is the sign, next 3 bits are the exponent and the last 4 bits are the matissa.

Thank you, have a good day.

Aristolochia
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2 Answers2

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  • convert the mantissa with the smallest exponent to the base of the largest exponent
  • process integer addition of mantissas (accounting for the sign)
  • possibly adjust the exponent so as to renormalize the resulting mantissa
Fabrice NEYRET
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First, in this form of binary representation of a floating number, the exponent is typically offset by a pre-agreed amount. For simplicity, let's say the offset is $(100)_2$. We can expand the two bit patterns as two numbers. $$ \require{enclose} \def\xD{{}_{10}} \def\xH{{}_{16}} \def\xB{{}_{2}} \newcommand{\xP}[2][black]{\enclose{box}{\color{#1}{\small\verb/#2/}}} \begin{array}{rcl} \xP[red]{1}\xP[orange]{100}\xP[green]{1100} &+& \xP[black]{0}\xP[blue]{101}\xP[magenta]{1011}\\ &\Downarrow\\ \color{red}{-}(1.\color{green}{1100})\xB \times 2^{(\color{orange}{100}) \xB-(100)\xB} &+& (1.\color{magenta}{1011})\xB \times 2^{(\color{blue}{101})\xB - (100)\xB} \\ &\Downarrow\\ \underbrace{-\left(1+\frac12+\frac14\right) \times 2^0 }_{= -\frac74} &+& \underbrace{\left(1 + \frac12 + \frac18 + \frac{1}{16}\right)\times 2^1}_{= \frac{27}{8}} \end{array} $$ The bit pattern on the left corresponds to the number $-\frac{7}{4}$, while the one on the right corresponds to $\frac{27}{8}$. Their sum equals to $\frac{13}{8}$. Since $$\frac{13}{8} = (1 + \frac12 + \frac18) \times 2^0 = (1.\color{gold}{1010})\xB \times 2^{(\color{orange}{100})\xB - (100)\xB}$$ The bit pattern for the result is $\xP{0}\xP[orange]{100}\xP[gold]{1010}$.

You can start from a different offset. However, the two numbers at hand and their sum all have compatible magnitude. There is no issues of underflow or overflow. Independent of what offset you choose, you will get the same final bit pattern.

achille hui
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  • What happens if I get the result 3/4? Do i just have to write the mantissa 1000 and the exponent 011? Thank you so much by the way – Aristolochia Nov 02 '15 at 02:06
  • @Aristolochia Yes. In the floating representation, there is a implied bit $1$ in front of the mantissa. If you have $\frac34$, you first rewrite it as $( 1 + \frac12 ) \times 2^{-1}$. The mantissa becomes $1000$ and the exponent becomes $(100)_2 - 1 = (011)_2$ as you suspected. – achille hui Nov 02 '15 at 04:09