C++ Institute Updated CLA-11-03 Exam Questions and Answers by sonia

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

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 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 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."