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

The program is a valid C program that can be compiled and run without errors. The program uses the argc and argv parameters of the main function, which are used to pass command-line arguments to the program. The argc parameter is an integer that stores the number of arguments, including the name of the program itself. The argv parameter is an array of strings that contains the arguments. The first element of the array, argv[0], is always the name of the program. The program declares an integer variable i and assigns it the value of 0. Then it prints the value of argv[i] as a string using the %s format specifier. Since i is 0, this is equivalent to printing argv[0], which is the name of the program. Therefore, the program outputs a predictable non-empty string, which is the name of the program. The exact name of the program may vary depending on how it is compiled and executed, but it will not be an empty string, an unpredictable string, or cause an execution failure. References = Command Line Arguments in C - GeeksforGeeks, Program Arguments (The GNU C Library), c - How to write a "argv" and "argc" - Stack Overflow

The program is not a valid C program and cannot be compiled successfully. The reason is that the program has a syntax error: the sizeof operator expects an expression or a type name as its operand, but the program uses s.s, which is not a valid member of the structure str. The structure str has three members: i, j, and k, but not s. Therefore, the compiler will report an error and the program will not run. References = sizeof operator in C - GeeksforGeeks, C Program to Find the Size of int, float, double and char, Sizeof operator in C - Online Tutorials Library

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.

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

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.

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 string literal "John" " " "Bean" is effectively concatenated into a single string by the compiler during compilation. Therefore, the value of p becomes a pointer to the string "John Bean". The printf statement then prints the string enclosed within square brackets, resulting in the output [John Bean].

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 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 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.