Pass Exam Questions Efficiently With CLA-11-03 Questions (2024)
CLA-11-03 Questions - Truly Beneficial For Your C++ Institute Exam
NEW QUESTION # 20
What happens if you try to compile and run this program?
#include <stdio.h>
int *fun(int *t) {
return t + 4;
}
int main (void) {
int arr[] = { 4, 3, 2, 1, 0 };
int *ptr;
ptr = fun (arr - 3);
printf("%d \n", ptr[2]);
return 0;
}
Choose the right answer:
- A. The program outputs 4
- B. The program outputs 3
- C. The program outputs 5
- D. The program outputs 2
- E. The program outputs 1
Answer: E
Explanation:
1.A function fun is defined that takes a pointer to an integer t and returns t + 4.
2.The main function defines an array arr with the elements { 4, 3, 2, 1, 0 }.
3.It then calls fun with the argument arr - 3. Since arr points to the first element of the array, arr - 3 is actually pointing to 3 positions before the start of the array, which is out of bounds.
4.Inside fun, t + 4 would effectively be arr + 1 (arr - 3 + 4).
5.The returned pointer from fun (which is arr + 1) is assigned to ptr.
6.ptr[2] is then the same as arr[1 + 2], which is arr[3]. The value at arr[3] is 1.
NEW QUESTION # 21
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 24
- C. The program outputs 16
- D. The program outputs 4
- E. The program outputs 8
Answer: D
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 # 22
What happens if you try to compile and run this program?
#include <stdio.h>
int main (int argc, char *argv[]) {
char *p = "John" " " "Bean";
printf("[%s]", p) ;
return 0;
}
Choose the right answer:
- A. The program outputs "[]"
- B. The program outputs nothing
- C. The program outputs [John Bean]
- D. The program outputs two lines of text
- E. The program outputs three lines of text
Answer: C
Explanation:
The string literal "John" " " "Bean" is effectively concatenated into a single string by the compiler during compilation. Therefore, the value of p becomes a pointer to the string "John Bean". The printf statement then prints the string enclosed within square brackets, resulting in the output [John Bean].
NEW QUESTION # 23
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. The program outputs a predictable non-empty string
- B. Compilation fails
- C. Execution fails
- D. The program outputs an unpredictable string, or execution fails
- E. The program outputs an empty string
Answer: A
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 # 24
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. Execution fails
- B. Compilation fails
- C. The program outputs ["]
- D. The program outputs []
- E. The program outputs []
Answer: A
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 # 25
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/=2)
if(i)
break ;
printf("%d",i);
return 0;
}
Choose the right answer:
The program executes an infinite loop
- A. Compilation fails
- B. The program outputs 0.5
- C. The program outputs 0
- D. The program outputs 1
Answer: D
Explanation:
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
NEW QUESTION # 26
What happens if you try to compile and run this program?
#include <stdio.h>
int main (int argc, char *argv[]) {
float f = 1e1 + 2e0 + 3e-1;
printf("%f ",f);
return 0;
}
Choose the right answer:
- A. The program outputs 123.00000
- B. Compilation fails
- C. The program outputs 1230.0000
- D. The program outputs 12.300000
- E. The program outputs 12300.000
Answer: D
Explanation:
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.
NEW QUESTION # 27
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. Execution fails
- C. The program outputs 4
- D. The program outputs 3
- E. The program outputs 2
Answer: D
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 # 28
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. Execution fails
- B. The program outputs 12
- C. Compilation fails
- D. The program outputs 16
- E. The program outputs 4
Answer: C
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 # 29
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. The program outputs 7
- B. Compilation fails
- C. The progham outputs 9
- D. The program outputs 10
- E. The program outputs 8
Answer: C
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 # 30
What happens if you try to compile and run this program?
#include <stdio.h>
int main (int argc, char *argv[]) {
int i = 20;
printf("%x", i);
return 0;
}
-
Choose the right answer:
- A. The program outputs 20
- B. Compilation fails
- C. The program outputs 24
- D. The program outputs 10
- E. The program outputs 14
Answer: E
Explanation:
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
NEW QUESTION # 31
What happens if you try to compile and run this program?
#include <stdio.h>
int fun(int i) {
return i++;
}
int main (void) {
int i = 1;
i = fun(i);
printf("%d",i);
return 0;
}
Choose the correct answer:
- A. Compilation fails
- B. The program outputs an unpredictable value
- C. The program outputs 0
- D. The program outputs 2
- E. The program outputs 1
Answer: E
Explanation:
In the fun function:
cCopy code
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:
cCopy code
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.
NEW QUESTION # 32
What happens if you try to compile and run this program?
#include <stdio.h>
#include <stdlib.h>
void fun (void) {
return 3.1415;
}
int main (int argc, char *argv[]) {
int i = fun(3.1415);
printf("%d",i);
return 0;
}
Choose the right answer:
- A. Execution fails
- B. Compilation fails
- C. The program outputs 3.1415
- D. The program outputs 4
- E. The program outputs 3
Answer: B
Explanation:
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 <stdio.h>
#include <stdlib.h>
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
NEW QUESTION # 33
What happens if you try to compile and run this program?
#include <stdio.h>
#include <string.h>
struct STR {
int i;
char c[20];
float f;
};
int main (int argc, char *argv[]) {
struct STR str = { 1, "Hello", 3 };
printf("%d", str.i + strlen(str.c));
return 0;
}
Choose the right answer:
- A. Compilation fails
- B. The program outputs 6
- C. The program outputs 4
- D. The program outputs 5
- E. The program outputs 1
Answer: B
Explanation:
The program defines a structure named STR that contains three members: an int, a char array, and a float.
Then it creates a variable of type struct STR named str and initializes its members with the values 1, "Hello", and 3. The program then prints the value of str.i + strlen(str.c), which is the sum of the int member and the length of the char array member. The length of the string "Hello" is 5, so the expression evaluates to 1 + 5 = 6.
Therefore, the program outputs 6. References = C struct (Structures) - Programiz, C Structures (structs) - W3Schools, C Structures - GeeksforGeeks
NEW QUESTION # 34
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 2
- B. The program outputs 0
- C. The program outputs 4
- D. The program outputs 3
- E. The program outputs 1
Answer: A
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 # 35
What happens if you try to compile and run this program?
#include <stdio.h>
int i = 0;
int main (int argc, char *argv[]) {
for(i; 1; i++);
printf("%d", i);
return 0;
}
Choose the right answer:
- A. Compilation fails
- B. The program executes an infinite loop
- C. The program outputs 0
- D. The program outputs 2
- E. The program outputs 1
Answer: B
Explanation:
The for loop in the program is initialized with i (which is 0), has the condition 1 (which is always true), and increments i in each iteration. Since the loop con-dition is always true, the loop will continue indefinitely, and i will keep incre-menting. The program will not reach the printf statement, and it will be stuck in an infinite loop.
*The program defines a global variable i and assigns it the value 0.
*The program defines a main function that takes two parameters: argc and argv.
*The program uses a for loop to increment the value of i as long as the condi-tion 1 is true, which is always the case.
*The program never exits the for loop, so it never reaches the printf function or the return statement.
*The program keeps running indefinitely, consuming CPU resources and memory. This is an example of a logical error in the program.
NEW QUESTION # 36
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