Solved Homework: Question: Goal: Names, Bindings and Scope 1. The scopes created by blocks, which could be nested in larger …

Goal: Names, Bindings and Scope 1. The scopes created by blocks, which could be nested in larger blocks are treated exactly those created by nested subprograms. C based languages Program Main; start Main ) do not have nested subprograms but they do have nested blocks. Create two nested blocks and declare variables of the same name. Assign values to those variables and print their values before you enter the block in the block and after you exit the block. Try it in Java and write down the difference, if there is any var r,v,a: integer procedure AREA: start AREA) begin a3 *r*r; endend AREA) procedure VOL3D start VOL3D) var a,r: integer; procedure MYPRINT: start MYPRINT.) begin println (Volume is,v) println (Area is , a) Define static, stack-dynamic, explicit-heap dynamic and implicit-heap dynamic variables. Then, please write example lines of codes in C to represent each variable, if possible 2. endend MYPRINT) begin a1 AREA v - 4/3*r*a; MYPRINT: endend VOL3D) 3. Examine the following program. Assume static scoping is used. List all the variables that are visible begin at points (1), (2), (3), and 4) (referencing environment) and specify where they are declared Write down what gets printed. If dynamic scoping is used what would get printed. Also give the end. l end Main h referencing environment for dynamic scoping as well a0: VOL3D

Goal: Names, Bindings and Scope 1. The scopes created by blocks, which could be nested in larger blocks are treated exactly those created by nested subprograms. C based languages Program Main; start Main ) do not have nested subprograms but they do have nested blocks. Create two nested blocks and declare variables of the same name. Assign values to those variables and print their values before you enter the block in the block and after you exit the block. Try it in Java and write down the difference, if there is any var r,v,a: integer procedure AREA: start AREA) begin a3 *r*r; endend AREA) procedure VOL3D start VOL3D) var a,r: integer; procedure MYPRINT: start MYPRINT.) begin println (“Volume is”,v) println (“Area is “, a) Define static, stack-dynamic, explicit-heap dynamic and implicit-heap dynamic variables. Then, please write example lines of codes in C to represent each variable, if possible 2. endend MYPRINT) begin a1 AREA v – 4/3*r*a; MYPRINT: endend VOL3D) 3. Examine the following program. Assume static scoping is used. List all the variables that are visible begin at points (1), (2), (3), and 4) (referencing environment) and specify where they are declared Write down what gets printed. If dynamic scoping is used what would get printed. Also give the end. l end Main h referencing environment for dynamic scoping as well a0: VOL3D

Expert Answer

Answer for Question 1 :

Nested blocks for C program will look as below :

#include <stdio.h> EASPONistsbinDebugVists.exe int main 0 Before Entering :5 Inside the Block:6 int ni n=5; / Declaring n outside After Exiting 5 Process returned 18 (ex12) execution time: .e50 s Press any key to continue. printf(Before Entering : %dn,n); int n = 6; printf(Inside Declaring n inside %dn,n); the Block : printf (After Exiting : %dn,n);

While for java it is as below :

But the above program gives error

Java code gives error because variable shadowing is not allowed in java while it is allowed in c that is main difference while scopes are considered in Java and C

While in C code the outside variable is shadowed if any other variable is declared inside and the scope of the inside variable finishes at the end of the block.

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