CLA-11-03 Sample Questions Answers

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 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 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 4 because the expression i << 2 performs a left shift operation on the binary representation of i, which is 00000010, by two bits, resulting in 00001000, which is equivalent to 8 in decimal. Then, the expression d /= 2 performs a division assignment operation, which divides d by 2 and assigns the result back to d, resulting in 4. The printf function then prints the value of d as a decimal integer using the %d format specifier.

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

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

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

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 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.