1 1 vote The minimum number of temporary variables to convert it into SSA is _________. a + b × c + d – e – a + b × c Assume order of precedence from highest to lowest as: ×, + and –. Consider associativity for + and × are not important but – is left associative. My ans : 5 Compiler Design static-single-assignment + – Anjan 1.1k views answer comment Share Follow Print See all 7 Comments 7 7 Comments reply Ashwin Kulkarni commented Jan 25, 2018 reply Follow flag Is it 6 ?? 0 0 replyShare Akshay Koli 4 commented Jan 25, 2018 reply Follow flag i'm also getting 6? 0 0 replyShare Anjan commented Jan 25, 2018 reply Follow flag t1=b*c; t2=a+t1; t3=t2+d; t4=t3-e; t5=t4-t2; im getting 5 ... 0 0 replyShare Ashwin Kulkarni commented Jan 25, 2018 reply Follow flag Look at that line "-" is left associative so you have to do d-e and then there result -a and then remaining things. 0 0 replyShare Mk Utkarsh commented Jan 25, 2018 reply Follow flag yeah same as Anjan 0 0 replyShare Anjan commented Jan 25, 2018 reply Follow flag I think left associative mean leftmost - is evaluated first. but not as "so you have to do d-e and then there result -a" please correct me if i'm wrong Ashwin Kulkarni 0 0 replyShare Shubhanshu commented Jan 25, 2018 reply Follow flag I am getting 5. $t1 = b*c$ $t2 = a + t1$ $t3 = t2 + d$ $t4 = t3-e$ $t5 = t4 - t2$ Paranthesization of the expression is:- $((((a+(b*c)+d)-e)-(a+(b*c)))$ Precedence of Operator is * > + > - 1 1 replyShare Please log in or register to add a comment.
