C++ Institute CLA-11-03 CLA - C Certified Associate Programmer Exam Practice Test

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 is a valid C program that can be compiled and run without errors.

•The variable i is declared as a char, which is an 8-bit signed integer type that can store val-ues from -128 to 127.

•The expression 20 + 020 + 0x20 evaluates to 68, because:

o20 is a decimal literal with the value 20

o020 is an octal literal with the value 16 (8^1 * 2 + 8^0 * 0)

o0x20 is a hexadecimal literal with the value 32 (16^1 * 2 + 16^0 * 0)

oThe + operator performs arithmetic addition on the operands and returns the sum

•The printf function prints the value of i as a decimal integer using the %d format specifier.

•The output of the program is 68.

To understand the behavior of this program, let's first analyze its structure:

1.It includes the standard I/O library, , and the standard library, , although is not used in this program.

2.main function declares a double variable x initialized to 1234567890.0.

3.It then prints x using %f format specifier in printf.

Key Points to Consider:

•The double data type in C is typically capable of representing a wide range of deci-mal numbers quite accurately.

•The %f format specifier in printf is used for outputting a float or double as a fixed-point number.

•There may be some precision issues when dealing with floating-point numbers, but these generally occur with more complex calculations or when the numbers are ex-tremely large or small.

Given that 1234567890.0 is a straightforward decimal number well within the representable range of a double, and the program doesn't perform any complex calculations, we can ex-pect the output to be quite close to the actual value of x.

However, due to the way floating-point numbers are represented and handled in C, there can be slight discrepancies in the least significant digits due to rounding or representation errors.

The program outputs 3 because the fun function returns the value of --n, which is a post-increment operator. This means that the value of n is decremented by 1 before it is returned. Therefore, fun() returns 3, which is the original value of n before decrementing. The main function calls fun() twice and adds the results, which gives 3 + 3 = 6. Then, the main function prints the result with a %d format specifier, which shows the integer part of the result. Therefore, the output of the program is:

fun() = 3 fun() = 3 printf(“%d \n”, fun() + fun()) = 6 = 3

The compilation fails because the program contains a syntax error. The identifier 1 is not a valid name for a variable, as it starts with a digit. Variable names in C must start with a letter or an under-score, and can contain letters, digits, or underscores. The compiler will report an error message such as error: expected identifier or '(' before numeric constant.

References = CLA – C Certified Associate Programmer Certification, C Essentials 1 - (Basics), C Varia-bles

The program defines a function fun with a static variable i. The main function declares two variables k and 1 (Note: The second variable has an invalid name, it should be changed to a valid identifier).

The fun function is called twice, and each time it increments the static variable i by 1. The values assigned to k and 1 become 2 and 3, respectively. The printf statement then prints the result of 1 + k, which is 3 + 2 equal to 5.

Therefore, the correct answer is "The program outputs 5."

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 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 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

1.A function fun is defined that takes a pointer to an integer t and returns t + 4.

2.The main function defines an array arr with the elements { 4, 3, 2, 1, 0 }.

3.It then calls fun with the argument arr - 3. Since arr points to the first element of the array, arr - 3 is actually pointing to 3 positions before the start of the array, which is out of bounds.

4.Inside fun, t + 4 would effectively be arr + 1 (arr - 3 + 4).

5.The returned pointer from fun (which is arr + 1) is assigned to ptr.

6.ptr[2] is then the same as arr[1 + 2], which is arr[3]. The value at arr[3] is 1.

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.