C++ Institute CLA - C Certified Associate Programmer CLA-11-03 Exam Dumps: Updated Questions & Answers (October 2025)

•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 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

In the f1 function, n = n * n; squares the input n and assigns the result back to n.

In the f2 function, f1(n) * f1(n) calls f1 twice with the input n and multiplies the results.

In the main function, printf("%d \n", f2(1)); prints the result of f2(1).

Let's calculate:

1.f1(1) returns 1 * 1 = 1.

2.f2(1) calls f1 twice with the input 1, so it's f1(1) * f1(1) = 1 * 1 * 1 * 1 = 1.

The program outputs 12.300000 because the printf function prints the value of f with a precision of 6 decimal places, which is the default precision for floating-point literals in C. The %f format specifier indicates that the argument is a floating-point value, and the space before it indicates that there should be a decimal point. The argument f is a float literal that represents 1e1 + 2e0 + 3e-1, which is equivalent to 1000000000 + 20000000 + 0.003 in decimal notation. Therefore, the output of the pro-gram is:

1e1 + 2e0 + 3e-1 = 1000000000 + 20000000 + 0.003 = 1230000000.003 = 123300000

The other options are incorrect because they either do not match the output of the program or do not use the correct format specifier for floating-point literals.

The program outputs 545454 because the fputs function writes the string “545454” to the file “file1” in binary mode, and the fscanf function reads the string as an integer from the file in text mode. The binary mode and the text mode are different ways of interpreting the data in a file. In binary mode, the data is stored as a sequence of bytes, and no translation is performed. In text mode, the data is stored as a sequence of characters, and some characters may be translated depending on the plat-form. For example, the newline character may be translated to a carriage return and a line feed on Windows, or just a line feed on Linux. The fopen function takes a mode argument that specifies whether the file should be opened in binary or text mode. The mode “wb” means write binary, and the mode “rt” means read text.

When the fputs function writes the string “545454” to the file in binary mode, it writes the ASCII val-ues of each character as a byte. The ASCII value of ‘5’ is 53, and the ASCII value of ‘4’ is 52. There-fore, the file contains the following bytes: 53 53 53 52 52 52. When the fscanf function reads the file in text mode, it interprets the bytes as characters and converts them to an integer using the %d format specifier. The %d format specifier expects a decimal integer, which is a number com-posed of digits from 0 to 9. Since the file contains only digits, the fscanf function successfully con-verts the string “545454” to an integer with the same value. 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 File I/O, ASCII Table

The program outputs 9 because the expression B + D + F evaluates to 9 using the enumeration constants defined by the enum keyword. The enum keyword creates a user-defined data type that can have one of a set of named values. By default, the first value is assigned 0, and each subsequent val-ue is assigned one more than the previous one, unless explicitly specified. Therefore, in this pro-gram, A is 0, B is 1, C is 2, D is 3, E is 4, and F is 5. The printf function then prints the sum of B, D, and F, which is 1 + 3 + 5 = 9, as a decimal integer using the %d format specifier.

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

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.

The declaration float **p; means that p is a pointer to a pointer to a float. It is used to declare a pointer that can point to another pointer, and that pointer, in turn, can point to a float.