CLA-11-03 Premium PDF & Test Engine Files with 41 Questions & Answers
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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 = 2;
int d= i << 2;
d /= 2;
printf ("%d", d) ;
return 0;
}
Choose the right answer:
- A. The program outputs 0
- B. The program outputs 2
- C. Compilation fails
- 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 # 19
What happens if you try to compile and run this program?
#include <stdio.h>
int f1(int n) {
return n = n * n;
}
int f2(int n) {
return n = f1(n) * f1(n);
}
int main(int argc, char ** argv) {
printf ("%d \n", f2(1));
return 0;
}
-
Select the correct answer:
- A. The program outputs 4
- B. The program outputs 1
- C. The program outputs 2
- D. Execution fails
- E. The program outputs 8
Answer: B
Explanation:
In the f1 function, n = n * n; squares the input n and assigns the result back to n.
In the f2 function, f1(n) * f1(n) calls f1 twice with the input n and multiplies the results.
In the main function, printf("%d \n", f2(1)); prints the result of f2(1).
Let's calculate:
1.f1(1) returns 1 * 1 = 1.
2.f2(1) calls f1 twice with the input 1, so it's f1(1) * f1(1) = 1 * 1 * 1 * 1 = 1.
NEW QUESTION # 20
What happens if you try to compile and run this program?
#include <stdio.h>
int main (int argc, char *argv[]) {
int i = 1, j = 0;
int 1 = !i + !! j;
printf("%d", 1);
return 0;
}
Choose the right answer:
- A. The program outputs 0
- B. The program outputs 3
- C. Compilation fails
- D. The program outputs 2
- E. The program outputs 1
Answer: C
Explanation:
The compilation fails because the program contains a syntax error. The identifier 1 is not a valid name for a variable, as it starts with a digit. Variable names in C must start with a letter or an under-score, and can contain letters, digits, or underscores. The compiler will report an error message such as error: expected identifier or '(' before numeric constant.
References = CLA - C Certified Associate Programmer Certification, C Essentials 1 - (Basics), C Varia-bles
NEW QUESTION # 21
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. The program outputs 0
- B. The program outputs 0.5
- C. The program outputs 1
- D. Compilation fails
Answer: C
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 # 22
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. The program outputs ["]
- B. Execution fails
- C. The program outputs []
- D. The program outputs []
- E. Compilation fails
Answer: B
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 # 23
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 4
- B. The program outputs 5
- C. The program outputs 3
- D. The program outputs 2
- E. The program outputs 1
Answer: B
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 # 24
What is the meaning of the following declaration?
float ** p;
Choose the right answer:
- A. p is a float pointer to a float
- B. p is a pointer to a float
- C. The declaration is erroneous
- D. p is a pointer to a pointer to a float
- E. p is a pointer to a float pointer
Answer: D
Explanation:
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.
NEW QUESTION # 25
Select the proper form for the following declaration:
p is a pointer to an array containing 10 int values
Choose the right answer:
- A. int * (p) [10];
- B. int (*p) [10];
- C. The declaration is invalid and cannot be coded in C
- D. int *p[10];
- E. int (*)p[10];
Answer: B
Explanation:
This is the correct way to declare a pointer to an array of 10 int values. The parentheses are necessary to indicate that p is a pointer to an array, not an array of pointers. The base type of p is 'an array of 10 int values'.12 References = 1: Pointer to an Array | Array Pointer - GeeksforGeeks 2: What is a pointer to array, int (*ptr) [10], and how does it work? - Stack Overflow
NEW QUESTION # 26
What happens if you try to compile and run this program?
#include <stdio.h>
struct s {
int i;
};
void fun(struct S st) {
st.i --;
int main (void) {
int k;
struct $ str1 = { 2 };
fun (str1) ;
k =str1.i;
printf("%d", k);
return 0;
}
-
Choose the correct answer:
- A. The program outputs 0
- B. The program outputs 2
- C. The program outputs 3
- D. Compilation fails
- E. The program outputs 1
Answer: B
Explanation:
The provided C program defines a struct S with one member i, a function fun that takes a struct S as an argument and decrements its i member, and a main function that creates a struct S, calls fun with it, and then prints the value of i from the struct.
Let's go through the main points of the program:
1.struct S str1 = { 2 }; initializes a struct S with i set to 2.
2.fun(str1); is called with str1 as an argument. However, since str1 is passed by val-ue, a copy of str1 is made for the function. The fun function then decrements i within its own copy of str1, and this decrement does not affect the original str1 in main.
3.k = str1.i; assigns the value of str1.i (which remains 2 because fun operated on a copy) to k.
4.printf("%d",k); prints the value of k, which is 2.
Given this, the correct answer is:
D: The program outputs 2.
NEW QUESTION # 27
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. The program outputs 7
- B. The program outputs 0
- C. The program outputs -1
- D. The program outputs 1
- E. Compilation fails
Answer: D
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 # 28
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. The progham outputs 9
- C. Compilation fails
- D. The program outputs 8
- E. The program outputs 10
Answer: B
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 # 29
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 4
- B. The program outputs 12
- C. The program outputs 16
- D. Compilation fails
- E. Execution fails
Answer: D
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 # 30
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. The program outputs 4
- B. The program outputs 3.1415
- C. The program outputs 3
- D. Compilation fails
- E. Execution fails
Answer: D
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 # 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. The program outputs 0
- B. The program outputs an unpredictable value
- C. The program outputs 1
- D. The program outputs 2
- E. Compilation fails
Answer: C
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
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. The program outputs 24
- B. The program outputs 16
- C. Compilation fails
- 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 # 33
What happens if you try to compile and run this program?
#include <stdio.h>
int main(int argc, char *argv[]) {
int i = 10 - 2 / 5 * 10 / 2 - 1;
printf("%d",i);
return 0;
}
Choose the right answer:
- A. The program outputs 0
- B. The program outputs 4
- C. The program outputs 9
- D. The program outputs 15
- E. Compilation fails
Answer: C
Explanation:
The expression 10 - 2 / 5 * 10 / 2 - 1 is evaluated based on the standard precedence rules in C. Division and multiplication have higher precedence than addition and subtrac-tion, and they are evaluated from left to right:
1.2 / 5 evaluates to 0 (integer division).
2.0 * 10 evaluates to 0.
3.0 / 2 evaluates to 0.
4.10 - 0 - 1 evaluates to 9.
Therefore, the correct answer is "The program outputs 9."
NEW QUESTION # 34
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