C++ Institute CLA-11-03 Exam Dumps

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 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 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 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.'

The program outputs 4 because the expression 2 / 1 + 4 / 2 evaluates to 4 using the integer arithmetic rules in C. The division operator / performs integer division when both operands are inte-gers, which means it discards the fractional part of the result. Therefore, 2 / 1 is 2 and 4 / 2 is 2, and their sum is 4. The printf function then prints the value of i as a decimal integer using the %d format specifier.

Reference = CLA -- C Certified Associate Programmer Certification, C Essentials 2 - (Intermediate), C Operators