CS 536 Practice Final Exam Fall 2018 1. Assume that we add a conditional expression of the form (Expr1 ? Expr2 : Expr3) to CSX. Expr1 is an expression that returns a bool. If Expr1 is true, expression Expr2 is eval- uated, and its value is the value of the conditional expression. If Expr1 is false, expres- sion Expr3 is evaluated, and its value is the value of the conditional expression. Outline the changes that would be needed in a CSX type-checker and code generator to handle conditional expressions. Illustrate your answer using the following example: a = (i != 0 ? j/i : 0); 2. (a) In CSX we type check a call to method M by first type checking M’s declaration. Then the actual parameters in the call to M are type checked. Finally, the number, type and kind of parameters found in the call are compared with the number, type and kind of parameters specified in M’s declaration. Consider the following alternative. In a declaration of a method P, no types are given to P’s parameters; they are simply given names. For example, int P(a,b,c) { ... } When a call to P is found, the parameters in the call are type checked (as usual) and then these types are used as definitions of the types of P’s parameters. What changes are needed in your type checking of method declarations and calls to implement this change? (b) A potential difficulty in the approach suggested in part (a) is that different calls to P may differ in the types used for a particular parameter. How would you handle a non-unique type for a parameter in different calls to P?-2- ! 3. (a) Code for an if statement is generated on the assumption that the value of the con- trol expression will not be known until run-time. In some cases the value of the control expression is known at compile-time. The simplest such case if when the control expression is just the boolean literal true or false. What changes would you make in your CSX code generator for if statements to handle the special case of a control expression that is either the literal true or the literal false? (b) The special case handled in part (a) is uncommon. More common is the case in which the control expression is an identifier declared to be a boolean constant with a literal initializer. For example, const debug = true; ... if (debug) ... What changes are needed to your solution to part (a) to include the case of identifiers declared to be boolean constants? (c) It may occur that the control expression of an if statement contains boolean oper- ators (&, |, !) whose operands are all boolean literals or boolean constants with literal initializers. For example, const debug1 = true; const debug2 = false; ... if (debug1 | debug2) ... What changes are needed to your solution to part (b) to include the case of boolean operators whose operands are all boolean literals or boolean constants? 4. Assume we have a Java class class K { int a; int sum(){ int b=1; return a+b; } } and the call z = (new K()).sum(); Explain the run-time steps needed to call and execute sum() (parameter passing, frame manipulation, return address manipulation, etc.)-3- ! 5. (a) Consider the following context free grammar: S → Label id ( ) ; S → Label id = id ; Label → id : Label → λ Is this grammar LL(1)? Why? Is this grammar LALR(1)? Why? (b) Consider the following context free grammar: S → Label id ( ) ; S → Label id = id ; Label → intlit : Label → λ Is this grammar LL(1)? Why? Is this grammar LALR(1)? Why? (c) Consider the following context free grammar: S → Label id ( ) : S → Label id ( Arg ) ; Label → intlit : Label → λ Arg → id Arg → λ Is this grammar LL(1)? Why? Is this grammar LALR(1)? Why?
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