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; }

Correct Answer: data types

Explanation:

The data types are mandatory for declaring the variables in the function pointers.

Correct Answer: 252

Explanation:

In this program, We are dividing the values of a and b, printing it.

Correct Answer: The function declaration should be preceded by the keyword friendThe function definitions does not use either the keyword or the scope operator :: The functions that are declared with the keyword friend as friend functionThus, a friend function is an ordinary function or a member of another class

Correct Answer: C requires all the variables to be declared at the beginning of a scope but in c++ we can declare variables anywhere in the scope This makes the programmer easier to understand because the variables are declared in the context of their use

Correct Answer: stack

Explanation:

Referencee is stored in stack

Correct Answer: #define

Correct Answer: 2

Explanation:

There are two types of macros. They are object-like and function-like.

Correct Answer: To structure a program into logical units.

Explanation:

The main aim of the namespace is to understand the logical units of the program and to make the program so robust.

Correct Answer: namespaceid::operator

Correct Answer: h