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What is the output of following program? abstract class Sum { public abstract int sumOfTwo(int n1, int n2); public abstract int sumOfThree(int n1, int n2, int n3); public void disp( ){ System.out.println("Method of class Sum"); } } class DemoAbstract1 extends Sum { public int sumOfTwo(int num1, int num2) { return num1+num2; } public int sumOfThree(int num1, int num2, int num3) { return num1+num2+num3; } public static void main(String args[ ]){ Sum obj = new DemoAbstract1( ); System.out.println(obj.sumOfTwo(3, 7)); System.out.println(obj.sumOfThree(4, 3, 19)); obj.disp( ); } }
Simplify the value of $$\frac{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} \times {\text{0}}{\text{.3}} - {\text{3}} \times 0.9 \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}}}}{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} - 0.9 \times {\text{0}}{\text{.2}} - {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}} - 0.3 \times 0.9}} = ?$$
What is the output of this program ? #include
using namespace std;
int n(char, int);
int (*p) (char, int) = n;
int main()
{
(*p)('d', 9);
p(10, 9);
return 0;
}
int n(char c, int i)
{
cout << c << i;
return 0;
}
$$\frac{{38 \times 38 \times 38 + 34 \times 34 \times 34 + 28 \times 28 \times 28 - 38 \times 34 \times 84}}{{38 \times 38 + 34 \times 34 + 28 \times 28 - 38 \times 34 - 34 \times 28 - 38 \times 28}}$$            is equal to = ?
Comment on the following 2 C programs.
#include  //Program 1int main(){ int a; int b; int c;}#include  //Program 2int main(){ int a; { int b; } { int c; }}
Which of the following is an external variable in the following C code?
#include int func (int a){ int b; return b;}int main(){ int c; func (c);}int d;
What will be the final values of i and j in the following C code?
#include int x = 0;int f(){ if (x == 0) return x + 1; else return x - 1;}int g(){ return x++;}int main(){ int i = (f() + g()) | g(); //bitwise or int j = g() | (f() + g()); //bitwise or}
What will be the final values of i and j in the following C code?
#include int x = 0;int f(){ if (x == 0) return x + 1; else return x - 1;}int g(){ return x++;}int main(){ int i = (f() + g()) || g(); int j = g() || (f() + g());}
Consider the following C function in which size is the number of elements in the array E: int MyX(int *E, unsigned int size) {       int Y = 0;       int Z;       int i, j, k;       for(i = 0; i < size; i++) Y = Y + E[i];       for(i = 0; i < size; i++) for(j = i; j < size; j++) {       Z = 0;       for(k = i; k <= j; k++) Z = Z + E[k];       if (Z > Y) Y = Z; }       return Y; } The value returned by the function MyX is the
What is the correct syntax of declaring array of pointers of integers of size 10 in C++? ; b) int **arr = new int*; c) int *arr = new int; d) int *arr = new int*;