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

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 program is not a valid C program and cannot be compiled successfully. The reason is that the program uses the same name main for both a function and a variable, which is not allowed in C. The name main is a reserved keyword that denotes the entry point of the program, and it cannot be redefined or reused for any other purpose. Therefore, the compiler will report an error and the program will not run. References = C - main() function - Tutorialspoint, C Keywords - GeeksforGeeks, C Basic Syntax

The program outputs 9 because the expression B + D + F evaluates to 9 using the enumeration constants defined by the enum keyword. The enum keyword creates a user-defined data type that can have one of a set of named values. By default, the first value is assigned 0, and each subsequent val-ue is assigned one more than the previous one, unless explicitly specified. Therefore, in this pro-gram, A is 0, B is 1, C is 2, D is 3, E is 4, and F is 5. The printf function then prints the sum of B, D, and F, which is 1 + 3 + 5 = 9, as a decimal integer using the %d format specifier.

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

The program defines a union named uni with three members: i, j, and k. The members share the same memory location. The values are assigned to s.i, s.j, and s.k, but since they share the same memory, the value of s.i will overwrite the values of s.j and s.k.

So, s.i will be 3, and both s.j and s.k will be 3. Then, the expression s.k * (s.i - s.j) be-comes 3 * (3 - 3), which equals 0. The printf statement prints the result, and the pro-gram outputs 0.

The program is a valid C program that can be compiled and run without errors. The program defines a union type named uni that contains three int members: i, j, and k. Then it creates a variable of type uni named s and assigns values to its members. However, since a union can only hold one member value at a time, the last assignment (s.k = 1) overwrites the previous values of s.i and s.j. Therefore, all the members of s have the same value of 1. The program then prints the value of s.k * (s.i - s.j), which is 1 * (1 - 1) = 0. Therefore, the program outputs 0. References = C Unions - GeeksforGeeks, C Unions (With Examples) - Programiz, C - Unions - Online Tutorials Library

•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 program outputs 14 because the printf function prints the value of i as a hexadecimal integer using the %x format specifier. The hexadecimal system uses 16 symbols to represent numbers, from 0 to 9 and from A to F. Each symbol corresponds to a decimal value, for example, A is 10, B is 11, C is 12, and so on. To convert a decimal number to a hexadecimal number, we need to divide the number by 16 repeatedly and write down the remainder in reverse order. For example, to convert 20 to hexa-decimal, we do:

20 / 16 = 1, remainder 4 1 / 16 = 0, remainder 1

The hexadecimal number is 14, as we write the remainders from right to left.

References = CLA – C Certified Associate Programmer Certification, C Essentials 2 - (Intermediate), C printf and scanf functions, Hexadecimal number system

The program outputs 3 because the expression p[2] - p[-1] evaluates to 3 using the pointer arithmetic rules in C. The pointer t points to the first element of the string literal "abcdefgh", which is stored in a read-only memory location. The pointer p is initialized to t + 2, which means it points to the third element of the string, which is 'c'. Then, p is incremented twice, so it points to the fifth ele-ment of the string, which is 'e'. The subscript operator [] is equivalent to adding an offset to the pointer and dereferencing it, so p[2] is the same as *(p + 2), which is 'g', and p[-1] is the same as *(p - 1), which is 'd'. The printf function then prints the difference between the ASCII values of 'g' and 'd', which is 103 - 100 = 3, as a decimal integer using the %d format specifier.

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

The program outputs 1 because the for loop terminates when i becomes 0. The for loop has no initialization, condition, or increment expressions, so it will run indefinitely unless a break statement is executed. The loop body consists of a single if statement that checks if i is non-zero,and if so, breaks out of the loop. Otherwise, i is divided by 2 and assigned back to itself. Since i is an integer, the division will truncate any fractional part. Therefore, the loop will iterate until i becomes 0, which will happen after one iteration, as 1 / 2 = 0. The printf function then prints the value of i as a deci-mal integer using the %d format specifier.

References = CLA – C Certified Associate Programmer Certification, [C Essentials 2 - (Intermediate)], C For Loop, C If…Else Statement

This is because when you initialize u with some values, only one member of u will be as-signed a value at a time, and the rest will remain uninitialized. Therefore, when you print sizeof(u), it will show the size of the largest member, which is f in this case. Since f is 4 bytes long, sizeof(u) will be 4 bytes as well.

If you want to learn more about unions in C programming, you can check out these re-sources:

•C Unions - GeeksforGeeks

•C Unions (With Examples) - Programiz

•C - Unions - Online Tutorials Library

The program is not a valid C program and cannot be compiled successfully. The reason is that the program has two syntax errors:

•The function fun has a void return type, which means it cannot return any value. However, the function tries to return a floating-point value of 3.1415, which is incompatible with the re-turn type. This will cause a compilation error.

•The function main is defined inside the function fun, which is not allowed in C. A function cannot be nested inside another function. This will also cause a compilation error.

To fix these errors, the function fun should have a double return type, and the function main should be defined outside the function fun. For example:

#include

#include

double fun (void) { return 3.1415; }

int main (int argc, char *argv[]) { int i = fun(3.1415); printf(“%d”,i); return 0; }

References = C - Functions - Tutorialspoint, C - return Statement - Tutorialspoint, C Basic Syntax

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 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.