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- Question
which keyword is used to define the macros in c++?
Options- A. macro
- B. define
- C. #define
- D. none of these
- Correct Answer
- #define
- 1. 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; }
Options- A. d99
- B. d9d9
- C. d9
- D. compile time error Discuss
- 2. What are the mandatory part to present in function pointers?
Options- A. &
- B. retrun values
- C. data types
- D. none of these Discuss
- 3. What is the output of this program? #include using namespace std; struct sec { int a; char b; }; int main() { struct sec s ={25,50}; struct sec *ps =(struct sec *)&s; cout << ps->a << ps->b; return 0; }
Options- A. 252
- B. 253
- C. 254
- D. 262 Discuss
- 4. What is friend function? Discuss
- 5. How variable declaration in c++ differs that in c? Discuss
- 6. How many types of macros are there in c++?
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
- 7. What is the use of Namespace?
Options- A. To encapsulate the data
- B. To structure a program into logical units.
- C. Both a and b
- D. none of these Discuss
- 8. What is the general syntax for accessing the namespace variable?
Options- A. namespaceid::operator
- B. namespace,operator
- C. namespace#operator
- D. none of the mentioned Discuss
- 9. What is the user-defined header file extension in c++?
Options- A. cpp
- B. h
- C. hf
- D. none of these Discuss
- 10. What do you mean by inline function? Discuss
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Search Results
Correct Answer: d99
Explanation:
In this program, we have declared the values as integer instead of character, So it is printing as d99 but it will not arise an error.
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: 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
Correct Answer: An inline function is a function that is expanded inline when invokedie the compiler replaces the function call with the corresponding function code An inline function is a function that is expanded in line when it is invoked That is the compiler replaces the function call with the corresponding function code (similar to macro)
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