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We want to declare x, y and z as pointers of type int. The alias name given is: intpt The correct way to do this using the keyword typedef is:

We want to declare x, y and z as pointers of type int. The alias name given is: intpt The correct way to do this using the keyword typedef is: Correct Answer typedef int* intptr;<br>intptr x,y,z;

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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( ); } }
We want to create an alias name for an identifier of the type unsigned long. The alias name is: ul. The correct way to do this using the keyword typedef is ____________
We want to create an alias name for an identifier of the type unsigned long. The alias name is: ul. The correct way to do this using the keyword typedef is . . . . . . . .
Consider the following grammar (that admits a series of declarations, followed by expressions) and the associated syntax directed translation (SDT) actions, given as pseudo-code: P → D* E* D → int ID {record that ID.lexeme is of type int} D → bool ID { record that ID.lexeme is of type bool} E → E1 + E2 {check that E1.type = E2.type = int; set E.type := int} E → !E1 {check that E1.type = bool; set E.type := bool} E → ID {set E.type := int} With respect to the above grammar; which one of the following choices is correct?
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*;
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;
}
The keyword typedef cannot be used to give alias names to pointers.
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