CLA-11-03 CLA - C Certified Associate Programmer Questions and Answers

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.

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 is a valid C program that can be compiled and run without errors. The program uses the || operator to perform a logical OR operation on the values of 7 and 0, which are both integer literals. The logical OR operator returns 1 if either operand is non-zero, and 0 otherwise. The program assigns the result of this operation to the variable i, which is an integer. The program then prints the value of !!i using the printf function. The !! operator is a double negation, which converts any non-zero value to 1, and 0 to 0. Since i is 1, !!i is also 1. Therefore, the program outputs 1.

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.

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.

The expression 10 - 2 / 5 * 10 / 2 - 1 is evaluated based on the standard precedence rules in C. Division and multiplication have higher precedence than addition and subtrac-tion, and they are evaluated from left to right:

1.2 / 5 evaluates to 0 (integer division).

2.0 * 10 evaluates to 0.

3.0 / 2 evaluates to 0.

4.10 - 0 - 1 evaluates to 9.

Therefore, the correct answer is "The program outputs 9."

he expression a++ && b++ involves the logical AND (&&) operator. In C, the logical AND op-erator short-circuits, meaning that if the left operand (a++ in this case) is false, the right operand (b++) is not evaluated.

Initially, a is 0, and b is 1. The result of a++ is 0 (false), so b++ is not evaluated. The value of b remains 1. The printf statement then prints the value of b, which is 1.

Therefore, the correct answer is "The program outputs 1."

References = CLA – C Associate Programmer documents

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