Question

Consider the following pseudocode Xinteger procedure set x(n global integer) Xin procedure print x write integer(x) procedure first set x(1) print x procedure second xinteger set x(2) print x set x(e) first() print x second print x What does this program print if the language uses static scoping? What does it print with dynamic scoping? Why?

Answer 1

Solution Static scoping: It refers top level environment Consider the following program for given pseudocode include <iostream> using namespace std int x void set (int n) X-n void print (int x) { cout<<"\n"<<x; } void first () set (1) cout<<"\n"<<x; void second () { int x set (2) cout<<"\n"<<x; } int main ( set (0) first ) { cout<<"\n"<<x; second () cout<<"\n"<<x;

The given pseudocode gives the following output if it runs on static scoping 1, 1, 2, 2 Еxplanation: In main, the first)method calls and prints"1*", the next print statement the value of "x" remain same as "1". The second() method calls and prints "2", the next print statement the value of "x" remain same as "2" Dynanic scoping The given pseudocode gives the following output if it runs on dynamic scoping: 1, 1, 2,1 Explanation In main), the first() method calls and prints "1*", the next print statement the value of "x" remain same as"1". The second) method calls and prints "2", the next print statement the value of "x" unchanged