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

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 4 because the expression i << 2 performs a left shift operation on the binary representation of i, which is 00000010, by two bits, resulting in 00001000, which is equivalent to 8 in decimal. Then, the expression d /= 2 performs a division assignment operation, which divides d by 2 and assigns the result back to d, resulting in 4. The printf function then prints the value of d as a decimal integer using the %d format specifier.

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

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 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

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.

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

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 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.