Solution :
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2. Consider the following programs (using C's syntax): #include <stdio.h> int a- 1, b-2; int foo...
Assignment: Show the symbol table for the following C programs at the printf lines (a) using lexical scope and (b) using dynamic scope. What does the program print using each kind of scope rule? Program 1: const int b = 10; int foo() { int a = b + 10; return a; } int bar() { int b = 4; return foo(); } int main() { printf(“foo = %d\n”,foo()); printf(“bar = %d\n”,bar()); return 0; } Program 2: int a; void...
Consider the following program: # include <iostream> int x = 3, y = 5; void foo(void) { x = x + 2; y = y + 4; } void bar(void) { int x = 10; y = y + 3; foo( ); cout << x << endl; cout << y << endl; } void baz(void) { int y = 7; bar( ); cout << y << endl; } void main( ) { baz( ); } What output does this program...
Consider the following program written in C syntax int a , b , c ; // first declaration void g() { print(a,b,c); } int f(int a) // parameter declaration { int b; // second declaration b = a + 1; g(); // first call { int a; // third declaration int c; // fourth declaration c = b; a = b + c; g(); // second call } g(); // third call return a + b ; } int main()...
B) The following code is given. Determine the outputs if scoping is static and dynamic. static scope output dynamic scope output #include <stdio.h> void function1(int); void function(int); int function(void); intx=10; main() { functioni(x); function2x); void function1(int y) { int x=y+5; printf("x=%d in function1 \n",function()); void function(int y){ int x=y+10; printf("x=%d in function1 \n", function()); int function(){ return x;
1. What is the output of the following program? include <stdio.h> int wilma (int x) if (x<5) x = 7; return (x) int main (void) int x-1 x=wilma (x) ; printf ("%d", x); return (0) b)3 c) 4 d) 7 a) 1 e) none of these
Consider the following code: int a:=10 //global int b:=12 //global proc F a:= a-b proc P (M:proc) int a:=2 M() proc K int b:=3 P(F) K() //main program print(a) //built in function a- what does this code print if it uses dynamic scoping and deep binding? b- what does this code print if it uses dynamic scoping and shallow binding?
Consider the following Java classes: class A{ public int foo () { return 1; } public void message () { System.out.println( "A" + foo()); } } class B extends A { public int foo() {return 2; } } class C extends B { public void message () { System.out.println( "C" + foo()); } } (i) What are the outputs of the following code? (ii) What would be the outputs if Java used static dispatching rather than dynamic dispatching? B b...
PLease explain output of these two programs: 1. #include <stdio.h> typedef struct { char *name; int x, y; int h, w; } box; typedef struct { unsigned int baud : 5; unsigned int div2 : 1; unsigned int use_external_clock : 1; } flags; int main(int argc, char** argv){ printf("The size of box is %d bytes\n", sizeof(box)); printf("The size of flags is %d bytes\n", sizeof(flags)); return 0; } 2. #include <stdio.h> #include <string.h> /* define simple structure */ struct { unsigned...
Hi, I want to devise an experiment to verify that F# is using static scooping as opposed to dynamic scoping. here is a similar program written in C language: int n = 3; int f() { return n; } // f must refer to an external (free) variable int main() { int x = 2; printf("%d", f()); // prints 3 if static, 2 if dynamically scoped return 0; } However, there's no main in F#. Thank you.
please evaluate the following code. this is JAVA a. class Car { public int i = 3; public Car(int i) { this.i = i; } } ... Car x = new Car(7), y = new Car(5); x = y; y.i = 9; System.out.println(x.i); b. class Driver { public static void main(String[] args) { int[] x = {5, 2, 3, 6, 5}; int n = x.length; for (int j = n-2; j > 0; j--) x[j] = x[j-1]; for (int j...