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

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 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

In the fun function:

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int fun(int i) { return i++; }

The post-increment operator i++ returns the current value of i and then increments it. So, fun(i) will return the current value of i (which is 1) and then increment i to 2.

In the main function:

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int i = 1; i = fun(i); printf("%d", i);

Here, i is assigned the result of fun(i), which is 1. So, the program prints the value of i, which is 1.

Therefore, the correct answer is D. The program outputs 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.

The program outputs 545454 because the fputs function writes the string “545454” to the file “file1” in binary mode, and the fscanf function reads the string as an integer from the file in text mode. The binary mode and the text mode are different ways of interpreting the data in a file. In binary mode, the data is stored as a sequence of bytes, and no translation is performed. In text mode, the data is stored as a sequence of characters, and some characters may be translated depending on the plat-form. For example, the newline character may be translated to a carriage return and a line feed on Windows, or just a line feed on Linux. The fopen function takes a mode argument that specifies whether the file should be opened in binary or text mode. The mode “wb” means write binary, and the mode “rt” means read text.

When the fputs function writes the string “545454” to the file in binary mode, it writes the ASCII val-ues of each character as a byte. The ASCII value of ‘5’ is 53, and the ASCII value of ‘4’ is 52. There-fore, the file contains the following bytes: 53 53 53 52 52 52. When the fscanf function reads the file in text mode, it interprets the bytes as characters and converts them to an integer using the %d format specifier. The %d format specifier expects a decimal integer, which is a number com-posed of digits from 0 to 9. Since the file contains only digits, the fscanf function successfully con-verts the string “545454” to an integer with the same value. 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 File I/O, ASCII Table

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 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 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.

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 provided program has a few key points to consider:

1.fun is a function that uses a static variable i. This means i retains its value between function calls. It's initialized to 1 and then incremented by 2 each time fun is called.

2.The main function calls fun twice, assigning the results to k and l (though there's a typo in the variable name l, it should be l = fun();, not 1 = fun();).

Let's step through the code:

•First call to fun: i starts at 1, increments by 2, so i becomes 3. This value (3) is as-signed to k.

•Second call to fun: i is now 3, increments by 2 again, so i becomes 5. This value (5) is assigned to l.

Finally, the printf statement attempts to print l - k, which is 5 - 3, resulting in 2.

So, the correct answer is:

A. The program outputs 2.

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 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