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Consider the following grammar to generate binary fractions:

If the above grammar with semantic rules calculate

and each non-terminal has synthesized attribute ‘val’ to store its value. Then the missing semantic rules will be _______.

A

B.

C.

D.

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crop it then paste question is not clearly viewed

Reason::

Lets first understand the grammar .

1.It says that an binary fraction can be of the form decimal(.) followed by any number of binary bits.

2. Any binary bits can start with 0 followed by any number of binary bits or 1 followed by any number of binary bits.

3. Or it (B) can be a single bit 1 or 0.

4. Now if its 0 i.e 0.0 then the decimal value is $\frac{0}{2^{-1}}$ i.e. 0 .

5. And if its 1 i.e. 0.1 then the decimal value is $\frac{1}{2^{-1}}$  i.e. 1/2 . Therefore S3 :{B.val = 1/2 }

6. Now for S2 , its says that the 1st bit is 1 follwed by several number of binary bits.Also let those binary bits be    $b_0b_1b_2......b_n$. Therefore the given binary number becomes 1 followed by  $b_0b_1b_2......b_n$.Therefore , the decimal value is $\frac{1}{2^{-1}}+\frac{b_0}{2^{-2}}+\frac{b_1}{2^{-3}}+.......+\frac{b_n}{2^{-n}}$.

7. Now taking 1/2 common from the 2nd , 3rd........ terms we get   $\frac{1}{2^{-1}}+\frac{1}{2}\left ( \frac{b_0}{2^{-1}}+\frac{b_1}{2^{-2}}+.......+\frac{b_n}{2^{-(n-1)}} \right)$ .

8. If we look clearly , it is nothing but 1/2+B.val/2. Therefore , S2: {B0.val=B1.val/2+1/2}.

9. Similarly, we can say for S1 :{B0.val=B1.val/2}.

I hope it helps.

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