ME OTS 5

Question

Consider the following grammar with corresponding synthesized attributes:
F → ⋅L {F.val = L.val}
L → LB {L.len = L.len + 1} {L.val = L.val+2 ^(-L.len)

× B.val}
L → B {L.len = 1, L.val = B.val/2}
B → 0 {B.val = 0}
B → 1 {B.val = 1}
If “F.val” gives the value of the binary fraction generated by F in above grammar then, the value of F.val on input {0.110} is _______. (Upto 3 decimal places)

 

 

Solution : 

 

Doubt : 

B.val = 0 ( red) 

then how 0.875 ( yellow ) ?

wont it be : 

0.75 + 2^-3 * 0 = 0.75 ?

 

 

Comments

in question you SDT as "L.val = L.val+2–L.len× B.val" but in solution you have written it in different way 0.75 + 2^-3 * 0 which is correct?

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@Hemanth_13 

edited but u can refer this.

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It looks like you are right.. Its just binary to decimal conversion

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@Hemanth_13 

okay.

oh.. i didn't notice that ( bin to dec ). thnx