Please suggest me the approach to solve this kind of questions
should be D only.
But in the original SDT,
if it is not L-attributed, then S -> A sign // A.sign = Sign.sign, S.value = A.value
if it is L-attributed, then S -> sign A // A.sign = Sign.sign, S.value = A.value
@Shaik Masthan if it is not L-attributed then S-attributed then how does A got the value of sign
@Shaik Masthan @Shobhit Joshi
what is difference between C) and D) option?
@srestha The description of SDT says if the sign is $1\,i.e\,'+'$ we will take minimum of the two values, in $(C)$ we are taking maximum in $'+'$ and in $(D)$ we take minimum at position $B1$ as $B1$ came when the sign was $'+'$
in synthesis attributes definition
s -> C- is with negative sign
c-> c1,n we have to take minimum of (c1.value and value(n)) as given in L-attribute definition