1. What will be the output of the following pseudo-code program? (lex declares a lexically-scoped variable,whereas dyn declares a dynamically-scopedvariable.)//define b as a dynamically-scoped var.dyn b = 41;//define a as a dynamically-scoped var.dyn a = 26;//define c as a lexically-scoped var.lex c = 15;//define function f3f3() {dyn b = 64; //redeclare bdyn a = 47; //redeclare alex c = 2; //redeclare cf1();}//define function f2f2() {print c, a, b;}//define function f1f1() {dyn b = 94; //redeclare bdyn a = 62; //redeclare af2();}f3(); //initial callRemember to justify your answer. 25-pointsSolution:The output for the above program will be 15 47 94

program L;var n : char;procedure W;beginwriteln(n);end;procedure D;var n : char;beginn := ‘D’;W;end;beginn := ‘L’;W;D;end.What is the output of the program under:(a) Static scope(b) Dynamic scope.

Which of the following can change during the execution of a program:Question 4Answera.a variable's valueb.a variable's typec.a variable's scoped.a variable's visibilitye.all of the above can change

Discuss in detail about the Syntax Directed Definitions.

Which of the following correctly describes the scope of a variable declared inside a function?Global scopeLocal scope to the functionStatic scopeFile scope

Which of the following is not the application of stack?A parentheses balancing programTracking of local variables at run timeCompiler Syntax AnalyzerData Transfer between two asynchronous process