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NEW QUESTION # 12
What happens if you try to compile and run this program?
#include <stdio.h>
int main (int argc, char *argv[]) {
int i = 2;
int d= i << 2;
d /= 2;
printf ("%d", d) ;
return 0;
}
Choose the right answer:
- A. Compilation fails
- B. The program outputs 0
- C. The program outputs 2
- D. The program outputs 1
- E. The program outputs 4
Answer: E
Explanation:
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
NEW QUESTION # 13
What happens if you try to compile and run this program?
#include <stdio.h>
int main (int argc, char *argv[]) {
int i = 1;
for(;i > 128;i *= 2);
printf("%d", i) ;
return 0;
}
-
Choose the right answer:
- A. Compilation fails
- B. The program outputs 128
- C. The program outputs a value greater than 128
- D. The program outputs a value less than 128
- E. The program enters an infinite loop
Answer: D
Explanation:
The main function declares an integer i and initializes it to 1. Then, it enters a for loop with no initialization statement, a condition i > 128, and an iteration expression i *= 2. The loop will continue to execute as long as i is greater than 128, but since i starts at 1, the loop's condition is false right from the start, meaning the loop body never executes.
However, this looks like an oversight, because usually, with this kind of loop, the intention is to run the loop until the condition becomes false. If the condition were i < 128, i would double each iteration until it reached or exceeded 128.
Given the current condition i > 128, the loop does nothing, and printf will output the ini-tial value of i, which is 1.
NEW QUESTION # 14
-
What happens if you try to compile and run this program?
#include <stdio.h>
int *f();
int main (int argc, char *argv[]) {
int *p;
p = f();
printf("%d",*p);
return 0;
}
int *f() {
static v = 1;
return &v;
}
Choose the right answer:
- A. The program outputs 3
- B. Compilation fails
- C. The program outputs 0
- D. The program outputs 1
- E. The program outputs 2
Answer: D
Explanation:
The program outputs 1 because the static variable v is initialized to 1 inside the f function, and it is visible to the main function. The f function returns the address of v, which is a pointer to an int. The main function dereferences the pointer and assigns it to p, which is another pointer to an int. Then, the main function prints the value of *p, which is the same as dereferencing p again. Therefore, the output of the program is:
f() = &v p = f() printf("%d",*p) = &v = 1
The other options are incorrect because they either do not match the output of the program or do not use the correct concept of static variables.
NEW QUESTION # 15
What happens if you try to compile and run this program?
#include <stdio.h>
int main (int argc, char *argv[]) {
char *t = "abcdefgh";
char *p = t + 2;
int i;
p++;
p++;
printf("%d ", p[2] - p[-1]);
return 0;
}
Choose the right answer:
- A. Compilation fails
- B. The program outputs 3
- C. Execution fails
- D. The program outputs 2
- E. The program outputs 4
Answer: B
Explanation:
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
NEW QUESTION # 16
What happens if you try to compile and run this program?
#include <stdio.h>
int main (int argc, char *argv[]) {
char *p = "World";
int i = 2;
switch (p[i]) {
case 'W' :i++; break ;
case 'o' :i += 2; break ;
case 'r' :i += 3; break ;
case '1' :i += 4; break ;
case 'd' :i += 5; break ;
default :i += 4;
}
printf("%d", i);
return 0;
}
-
Choose the right answer:
- A. The program outputs 3
- B. Compilation fails
- C. The program outputs 6
- D. The program outputs 4
- E. The program outputs 5
Answer: E
Explanation:
*The program defines a pointer p that points to the string literal "World".
*The program also defines an integer variable i and assigns it the value 2.
*The program uses a switch statement to check the value of p[i], which is the third character of the string
"World", i.e. 'r'.
*The program finds a matching case for 'r' and executes the statement i += 3;, which adds 3 to the value of i.
Then it breaks out of the switch statement.
*The program prints the value of i, which is now 5, using the printf function.
*The program returns 0 and exits.
NEW QUESTION # 17
What happens when you compile and run the following program?
#include <stdio.h>
int fun(void) {
static int i = 1;
i++;
return i;
}
int main (void) {
int k, l;
k = fun ();
l = fun () ;
printf("%d",l + k);
return 0;
}
Choose the right answer:
- A. The program outputs 3
- B. The program outputs 2
- C. The program outputs 4
- D. The program outputs 5
- E. The program outputs 1
Answer: D
Explanation:
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."
NEW QUESTION # 18
What happens if you try to compile and run this program?
#include <stdio.h>
int main (int argc, char *argv[]) {
int i = 7 || 0 ;
printf("%d", !! i);
return 0;
}
Choose the right answer:
- A. Compilation fails
- B. The program outputs 0
- C. The program outputs 1
- D. The program outputs -1
- E. The program outputs 7
Answer: C
Explanation:
The program is a valid C program that can be compiled and run without errors. The program uses the || operator to perform a logical OR operation on the values of 7 and 0, which are both integer literals. The logical OR operator returns 1 if either operand is non-zero, and 0 otherwise. The program assigns the result of this operation to the variable i, which is an integer. The program then prints the value of !!i using the printf function. The !! operator is a double negation, which converts any non-zero value to 1, and 0 to 0. Since i is 1,
!!i is also 1. Therefore, the program outputs 1.
NEW QUESTION # 19
What happens if you try to compile and run this program?
#include <stdio.h>
int main (int argc, char *argv[]) {
int i = 0;
printf ("%s", argv[i]);
return 0;
}
Choose the right answer:
- A. Compilation fails
- B. Execution fails
- C. The program outputs a predictable non-empty string
- D. The program outputs an unpredictable string, or execution fails
- E. The program outputs an empty string
Answer: C
Explanation:
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
NEW QUESTION # 20
Assume that ints are 32-bit wide.
What happens if you try to compile and run this program?
#include <stdio.h>
typedef union {
int i;
int j;
int k;
} uni;
int main (int argc, char *argv[]) {
uni s;
s.i = 3;
s.j = 2;
s.k = 1;
printf("%d",s.k * (s.i - s.j));
return 0;
}
Choose the right answer:
- A. The program outputs 3
- B. The program outputs 9
- C. Compilation fails
- D. Execution fails
- E. The program outputs 0
Answer: E
Explanation:
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
NEW QUESTION # 21
What happens if you try to compile and run this program?
#include <stdio.h>
int main (int argc, char *argv[]) {
char *s = "\\\"\\\\";
printf ("[%c]", s [1]);
return 0;
}
Choose the right answer:
- A. Compilation fails
- B. The program outputs []
- C. The program outputs []
- D. Execution fails
- E. The program outputs ["]
Answer: D
Explanation:
In the program, the character array char *s = "\\\"\\\\"; is defined with the value "\"\\". When printing s[1] using printf("[%c]", s[1]);, it prints the character at index 1 of the string.
Here's the breakdown of the string \\\"\\\\:
*s[0] is '\'
*s[1] is '"'
So, the program outputs ["]. Therefore, the correct answer is B. The program outputs ["]
NEW QUESTION # 22
What happens if you try to compile and run this program?
#include <stdio.h>
#include <stdlib.h>
int main (int argc, char *argv[]) {
double x = 1234567890.0;
printf ("%f",x);
return 0;
}
Choose the most precise answer:
- A. The program outputs 1234567900.0
- B. Compilation fails
- C. The program outputs a value greater than 1234500000.0 and less than 1234600000.0
- D. Execution fails
- E. The program outputs 1234567890.0
Answer: E
Explanation:
To understand the behavior of this program, let's first analyze its structure:
1.It includes the standard I/O library, <stdio.h>, and the standard library, <stdlib.h>, although <stdlib.h> 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.
NEW QUESTION # 23
What happens when you compile and run the following program?
#include <stdio.h>
int fun (void) {
static int i = 1;
i += 2;
return i;
}
int main (void) {
int k, 1;
k = fun ();
1 = fun () ;
printf ("%d", 1 - k);
return 0;
}
Choose the right answer:
- A. The program outputs 3
- B. The program outputs 0
- C. The program outputs 2
- D. The program outputs 4
- E. The program outputs 1
Answer: C
Explanation:
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.
NEW QUESTION # 24
Assume that ints and floats are 32-bit wide.
What happens if you try to compile and run this program?
#include <stdio.h>
union uni {
float f, g;
int i, j;
};
int main (int argc, char *argv[]) {
union uni u;
printf ("%ld", sizeof (u) ) ;
return 0;
}
Choose the right answer:
- A. Compilation fails
- B. The program outputs 16
- C. The program outputs 24
- D. The program outputs 8
- E. The program outputs 4
Answer: E
Explanation:
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
NEW QUESTION # 25
Assume that ints are 32-bit wide.
What happens if you try to compile and run this program?
#include <stdio.h>
typedef struct
int i;
int j;
int k;
} str;
int main (int argc, char *argv[]) {
str s = { 7, 7, 7 };
printf ("%d", sizeof (s.s));
return 0;
}
Choose the right answer:
- A. The program outputs 12
- B. Execution fails
- C. The program outputs 16
- D. The program outputs 4
- E. Compilation fails
Answer: E
Explanation:
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
NEW QUESTION # 26
What happens if you try to compile and run this program?
#include <stdio.h>
int main (int argc, char *argv[]) {
int i =2, j = 1;
if(i / j)
j += j;
else
i += i;
printf("%d",i + j);
return 0;
}
Choose the right answer:
- A. The program outputs 3
- B. Compilation fails
- C. The program outputs 5
- D. The program outputs 1
- E. The program outputs 4
Answer: E
Explanation:
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.
NEW QUESTION # 27
What happens if you try to compile and run this program?
enum { A, B, C, D, E, F };
#include <stdio.h>
int main (int argc, char *argv[]) {
printf ("%d", B + D + F);
return 0;
}
Choose the right answer:
- A. Compilation fails
- B. The program outputs 10
- C. The program outputs 7
- D. The progham outputs 9
- E. The program outputs 8
Answer: D
Explanation:
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
NEW QUESTION # 28
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Free C++ Institute CLA-11-03 Exam 2024 Practice Materials Collection: https://www.practicematerial.com/CLA-11-03-exam-materials.html
CLA-11-03 Exam Info and Free Practice Test All-in-One Exam Guide May-2024: https://drive.google.com/open?id=139gCaGTzqkPpvtKhzF44wFT9OoBoOKln
