CLA-11-03 Exam Dumps - C++ Institute Certification Questions and Answers

The program outputs 4 because the expression 2 / 1 + 4 / 2 evaluates to 4 using the integer arithmetic rules in C. The division operator / performs integer division when both operands are inte-gers, which means it discards the fractional part of the result. Therefore, 2 / 1 is 2 and 4 / 2 is 2, and their sum is 4. The printf function then prints the value of i as a decimal integer using the %d format specifier.

References = CLA – C Certified Associate Programmer Certification, C Essentials 2 - (Intermediate), C Operators

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 main function declares an integer i and initializes it to 1. Then, it enters a for loop with no initialization statement, a condition i > 128, and an iteration expression i *= 2. The loop will continue to execute as long as i is greater than 128, but since i starts at 1, the loop's condition is false right from the start, meaning the loop body never executes.

However, this looks like an oversight, because usually, with this kind of loop, the intention is to run the loop until the condition becomes false. If the condition were i < 128, i would double each iteration until it reached or exceeded 128.

Given the current condition i > 128, the loop does nothing, and printf will output the ini-tial value of i, which is 1.

The program outputs 3 because the expression p[2] - p[-1] evaluates to 3 using the pointer arithmetic rules in C. The pointer t points to the first element of the string literal "abcdefgh", which is stored in a read-only memory location. The pointer p is initialized to t + 2, which means it points to the third element of the string, which is 'c'. Then, p is incremented twice, so it points to the fifth ele-ment of the string, which is 'e'. The subscript operator [] is equivalent to adding an offset to the pointer and dereferencing it, so p[2] is the same as *(p + 2), which is 'g', and p[-1] is the same as *(p - 1), which is 'd'. The printf function then prints the difference between the ASCII values of 'g' and 'd', which is 103 - 100 = 3, as a decimal integer using the %d format specifier.

References = CLA – C Certified Associate Programmer Certification, C Essentials 2 - (Intermediate), C Pointers, C Strings