CLA-11-03 C++ Institute CLA - C Certified Associate Programmer Free Practice Exam Questions (2025 Updated)

The provided program has a few key points to consider:

1.fun is a function that uses a static variable i. This means i retains its value between function calls. It's initialized to 1 and then incremented by 2 each time fun is called.

2.The main function calls fun twice, assigning the results to k and l (though there's a typo in the variable name l, it should be l = fun();, not 1 = fun();).

Let's step through the code:

•First call to fun: i starts at 1, increments by 2, so i becomes 3. This value (3) is as-signed to k.

•Second call to fun: i is now 3, increments by 2 again, so i becomes 5. This value (5) is assigned to l.

Finally, the printf statement attempts to print l - k, which is 5 - 3, resulting in 2.

So, the correct answer is:

A. The program outputs 2.

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 is a valid C program that can be compiled and run without errors. The program uses the argc and argv parameters of the main function, which are used to pass command-line arguments to the program. The argc parameter is an integer that stores the number of arguments, including the name of the program itself. The argv parameter is an array of strings that contains the arguments. The first element of the array, argv[0], is always the name of the program. The program declares an integer variable i and assigns it the value of 0. Then it prints the value of argv[i] as a string using the %s format specifier. Since i is 0, this is equivalent to printing argv[0], which is the name of the program. Therefore, the program outputs a predictable non-empty string, which is the name of the program. The exact name of the program may vary depending on how it is compiled and executed, but it will not be an empty string, an unpredictable string, or cause an execution failure. References = Command Line Arguments in C - GeeksforGeeks, Program Arguments (The GNU C Library), c - How to write a "argv" and "argc" - Stack Overflow

The string literal "John" " " "Bean" is effectively concatenated into a single string by the compiler during compilation. Therefore, the value of p becomes a pointer to the string "John Bean". The printf statement then prints the string enclosed within square brackets, resulting in the output [John Bean].

In the program, the character array char *s = "\\\"\\\\"; is defined with the value "\"\\". When printing s[1] using printf("[%c]", s[1]);, it prints the character at index 1 of the string.

Here's the breakdown of the string \\\"\\\\:

•s[0] is '\'

•s[1] is '"'

So, the program outputs ["]. Therefore, the correct answer is B. The program outputs ["]

The program outputs 1 because the for loop terminates when i becomes 0. The for loop has no initialization, condition, or increment expressions, so it will run indefinitely unless a break statement is executed. The loop body consists of a single if statement that checks if i is non-zero,and if so, breaks out of the loop. Otherwise, i is divided by 2 and assigned back to itself. Since i is an integer, the division will truncate any fractional part. Therefore, the loop will iterate until i becomes 0, which will happen after one iteration, as 1 / 2 = 0. The printf function then prints the value of i as a deci-mal integer using the %d format specifier.

References = CLA – C Certified Associate Programmer Certification, [C Essentials 2 - (Intermediate)], C For Loop, C If…Else Statement

Let's analyze the macros and the program:

1.ALPHA is defined as 0.

2.BETA is defined as ALPHA - 1, which is 0 - 1.

3.GAMMA is defined as 1.

4.DELTA is defined as ALPHA - BETA - GAMMA. With the previous definitions, this expands to 0 - (0 - 1) - 1.

Now, let's expand DELTA with the given values:

makefileCopy code

DELTA = 0 - (0 - 1) - 1 DELTA = 0 - 0 + 1 - 1 DELTA = 0 + 1 - 1 DELTA = 1 - 1 DELTA = 0

It is important to note that the macro dELTA is defined with a lowercase 'd', but the printf function is trying to print DELTA with an uppercase 'D'. Preprocessor tokens are case-sensitive, so this is a mismatch. However, for the sake of the question, let's assume that dELTA was meant to be DELTA with an uppercase 'D'.

Since the actual calculation results in 0, but there is a typo in the printf statement (it should print dELTA, not DELTA), the compilation will fail due to DELTA not being defined.

In the if statement, i / j is 2 / 1, which is true. Therefore, the if block is executed, and j += j; doubles the value of j (j becomes 2).

After the if-else statement, printf("%d", i + j); prints the sum of i and the updated val-ue of j (2 + 2), which is 4.

The program outputs 1 because the static variable v is initialized to 1 inside the f function, and it is visible to the main function. The f function returns the address of v, which is a pointer to an int. The main function dereferences the pointer and assigns it to p, which is another pointer to an int. Then, the main function prints the value of *p, which is the same as dereferencing p again. Therefore, the output of the program is:

f() = &v p = f() printf(“%d”,*p) = &v = 1

The other options are incorrect because they either do not match the output of the program or do not use the correct concept of static variables.

The program is not a valid C program and cannot be compiled successfully. The reason is that the program has two syntax errors:

•The function fun has a void return type, which means it cannot return any value. However, the function tries to return a floating-point value of 3.1415, which is incompatible with the re-turn type. This will cause a compilation error.

•The function main is defined inside the function fun, which is not allowed in C. A function cannot be nested inside another function. This will also cause a compilation error.

To fix these errors, the function fun should have a double return type, and the function main should be defined outside the function fun. For example:

#include

#include

double fun (void) { return 3.1415; }

int main (int argc, char *argv[]) { int i = fun(3.1415); printf(“%d”,i); return 0; }

References = C - Functions - Tutorialspoint, C - return Statement - Tutorialspoint, C Basic Syntax

The program defines a structure named STR that contains three members: an int, a char array, and a float. Then it creates a variable of type struct STR named str and initializes its members with the values 1, “Hello”, and 3. The program then prints the value of str.i + strlen(str.c), which is the sum of the int member and the length of the char array member. The length of the string “Hello” is 5, so the expression evaluates to 1 + 5 = 6. Therefore, the program outputs 6. References = C struct (Structures) - Programiz, C Structures (structs) - W3Schools, C Structures - GeeksforGeeks