CLA-11-03 Practice Exam Questions with Answers CLA - C Certified Associate Programmer Certification

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 provided C program defines a struct S with one member i, a function fun that takes a struct S as an argument and decrements its i member, and a main function that creates a struct S, calls fun with it, and then prints the value of i from the struct.

Let's go through the main points of the program:

1.struct S str1 = { 2 }; initializes a struct S with i set to 2.

2.fun(str1); is called with str1 as an argument. However, since str1 is passed by val-ue, a copy of str1 is made for the function. The fun function then decrements i within its own copy of str1, and this decrement does not affect the original str1 in main.

3.k = str1.i; assigns the value of str1.i (which remains 2 because fun operated on a copy) to k.

4.printf("%d",k); prints the value of k, which is 2.

Given this, the correct answer is:

D. The program outputs 2.

The for loop in the program is initialized with i (which is 0), has the condition 1 (which is always true), and increments i in each iteration. Since the loop con-dition is always true, the loop will continue indefinitely, and i will keep incre-menting. The program will not reach the printf statement, and it will be stuck in an infinite loop.

•The program defines a global variable i and assigns it the value 0.

•The program defines a main function that takes two parameters: argc and argv.

•The program uses a for loop to increment the value of i as long as the condi-tion 1 is true, which is always the case.

•The program never exits the for loop, so it never reaches the printf function or the return statement.

•The program keeps running indefinitely, consuming CPU resources and memory. This is an example of a logical error in the program.

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 600 because the #ifdef directive checks if the macro SYM is defined, and if so, executes the code between it and the corresponding #else or #endif directive. Otherwise, it skips that code and executes the code after the #else directive, if any. In this program, the macro SYM is defined by the #define directive, but then undefined by the #undef directive, which removes the def-inition of a macro. Therefore, the code between the #ifdef and the #else directives is skipped, and the code after the #else directive is executed, which assigns 200 to the variable i. The variable j is then assigned the sum of i and 200, which is 400. The printf function then prints the sum of i and j, which is 600, as a decimal integer using the %d format specifier.

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

The program is not a valid C program and cannot be compiled successfully. The reason is that the program uses the same name main for both a function and a variable, which is not allowed in C. The name main is a reserved keyword that denotes the entry point of the program, and it cannot be redefined or reused for any other purpose. Therefore, the compiler will report an error and the program will not run. References = C - main() function - Tutorialspoint, C Keywords - GeeksforGeeks, C Basic Syntax

The program outputs 14 because the printf function prints the value of i as a hexadecimal integer using the %x format specifier. The hexadecimal system uses 16 symbols to represent numbers, from 0 to 9 and from A to F. Each symbol corresponds to a decimal value, for example, A is 10, B is 11, C is 12, and so on. To convert a decimal number to a hexadecimal number, we need to divide the number by 16 repeatedly and write down the remainder in reverse order. For example, to convert 20 to hexa-decimal, we do:

20 / 16 = 1, remainder 4 1 / 16 = 0, remainder 1

The hexadecimal number is 14, as we write the remainders from right to left.

References = CLA – C Certified Associate Programmer Certification, C Essentials 2 - (Intermediate), C printf and scanf functions, Hexadecimal number system