[Apr 30, 2024] Latest Questions CLA-11-03 Guide to Prepare Free Practice Tests
Reliable CLA-11-03 Dumps Questions Available as Web-Based Practice Test Engine
NEW QUESTION # 18
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: E
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 # 19
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 -1
- B. The program outputs 0
- C. The program outputs 1
- D. Compilation fails
- 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 # 20
What happens if you try to compile and run this program?
#include <stdio.h>
int main(int argc, char *argv[]) {
int i = 2 / 1 + 4 / 2;
printf("%d",i);
return 0;
}
Choose the right answer:
- A. The program outputs 0
- B. The program outputs 3
- C. The program outputs 4
- D. The program outputs 5
- E. Compilation fails
Answer: C
Explanation:
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
NEW QUESTION # 21
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 3.1415
- B. The program outputs 3
- C. The program outputs 4
- D. Execution fails
- 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 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 # 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 [John Bean]
- B. The program outputs two lines of text
- C. The program outputs three lines of text
- D. The program outputs "[]"
- E. The program outputs nothing
Answer: A
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 = 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 4
- C. The program outputs 1
- D. Compilation fails
- E. The program outputs 2
Answer: B
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 # 24
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. The program outputs 128
- B. Compilation fails
- C. The program outputs a value greater than 128
- D. The program enters an infinite loop
- E. The program outputs a value less than 128
Answer: E
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 # 25
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 5
- B. The program outputs 3
- C. The program outputs 1
- D. The program outputs 4
- E. The program outputs 2
Answer: A
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 # 26
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. The program outputs 4
- B. Compilation fails
- C. Execution fails
- 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 # 27
What is the meaning of the following declaration?
float ** p;
Choose the right answer:
- A. The declaration is erroneous
- B. p is a float pointer to a float
- C. p is a pointer to a float pointer
- D. p is a pointer to a pointer to a float
- E. p is a pointer to a float
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 # 28
What happens when you compile and run the following program?
#include <stdio.h>
#define SYM
#define BOL 100
#undef SYM
int main (void) {
#ifdef SYM
int i = 100;
#else
int i= 200;
#endif
int j = i + 200;
printf("%d",i+j);
return 0;
}
Select the correct answer:
- A. The program outputs 300
- B. The program outputs 100
- C. The program outputs 200
- D. The program outputs 400
- E. The program outputs 600
Answer: E
Explanation:
The program outputs 600 because the #ifdef directive checks if the macro SYM is defined, and if so, executes the code between it and the corresponding #else or #endif directive. Otherwise, it skips that code and executes the code after the #else directive, if any. In this program, the macro SYM is defined by the #define directive, but then undefined by the #undef directive, which removes the def-inition of a macro. Therefore, the code between the #ifdef and the #else directives is skipped, and the code after the #else directive is executed, which assigns 200 to the variable i. The variable j is then assigned the sum of i and 200, which is 400. The printf function then prints the sum of i and j, which is 600, as a decimal integer using the %d format specifier.
References = CLA - C Certified Associate Programmer Certification, C Essentials 2 - (Intermediate), C Preprocessor
NEW QUESTION # 29
-
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 0
- B. The program outputs 1
- C. The program outputs 3
- D. Compilation fails
- E. The program outputs 2
Answer: B
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 # 30
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. The program outputs 0
- B. The program executes an infinite loop
- C. The program outputs 1
- D. Compilation fails
- E. The program outputs 2
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 # 31
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. The program outputs 6
- B. The program outputs 5
- C. The program outputs 1
- D. The program outputs 4
- E. Compilation fails
Answer: A
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 # 32
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 5
- B. The program outputs 6
- C. The program outputs 3
- D. The program outputs 4
- E. Compilation fails
Answer: A
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 # 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 15
- C. The program outputs 9
- D. The program outputs 4
- 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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