C++ Institute CLA-11-03 Dumps Questions Answers

•The program defines a pointer p that points to the string literal "World".

•The program also defines an integer variable i and assigns it the value 2.

•The program uses a switch statement to check the value of p[i], which is the third character of the string "World", i.e. 'r'.

•The program finds a matching case for 'r' and executes the statement i += 3;, which adds 3 to the value of i. Then it breaks out of the switch statement.

•The program prints the value of i, which is now 5, using the printf function.

•The program returns 0 and exits.

The program defines a union named uni with three members: i, j, and k. The members share the same memory location. The values are assigned to s.i, s.j, and s.k, but since they share the same memory, the value of s.i will overwrite the values of s.j and s.k.

So, s.i will be 3, and both s.j and s.k will be 3. Then, the expression s.k * (s.i - s.j) be-comes 3 * (3 - 3), which equals 0. The printf statement prints the result, and the pro-gram outputs 0.

The program is a valid C program that can be compiled and run without errors. The program defines a union type named uni that contains three int members: i, j, and k. Then it creates a variable of type uni named s and assigns values to its members. However, since a union can only hold one member value at a time, the last assignment (s.k = 1) overwrites the previous values of s.i and s.j. Therefore, all the members of s have the same value of 1. The program then prints the value of s.k * (s.i - s.j), which is 1 * (1 - 1) = 0. Therefore, the program outputs 0. References = C Unions - GeeksforGeeks, C Unions (With Examples) - Programiz, C - Unions - Online Tutorials Library

The program outputs 3 because the fun function returns the value of --n, which is a post-increment operator. This means that the value of n is decremented by 1 before it is returned. Therefore, fun() returns 3, which is the original value of n before decrementing. The main function calls fun() twice and adds the results, which gives 3 + 3 = 6. Then, the main function prints the result with a %d format specifier, which shows the integer part of the result. Therefore, the output of the program is:

fun() = 3 fun() = 3 printf(“%d \n”, fun() + fun()) = 6 = 3