100% Free CLA-11-03 Exam Dumps Use Real C++ Institute Certification Dumps With 41 Questions!
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NEW QUESTION # 17
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. The program outputs 1
- C. The program outputs 4
- D. Compilation fails
- E. The program outputs 5
Answer: C
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 # 18
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 3
- B. The program outputs 1
- C. Compilation fails
- D. The program outputs 2
- E. The program outputs 0
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 # 19
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 1
- B. Compilation fails
- C. The program outputs 2
- D. The program executes an infinite loop
- E. The program outputs 0
Answer: D
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 # 20
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 an unpredictable value
- B. Compilation fails
- C. The program outputs 2
- D. The program outputs 0
- 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 # 21
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: A
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 # 22
What happens if you try to compile and run this program?
#include <stdio.h>
fun (void) {
static int n = 3;
return --n;
}
int main (int argc, char ** argv) {
printf("%d \n", fun() + fun());
return 0;
}
Select the correct answer:
- A. The program outputs 1
- B. The program outputs 3
- C. The program outputs 4
- D. The program outputs 2
- E. The program outputs 0
Answer: B
Explanation:
The program outputs 3 because the fun function returns the value of --n, which is a post-increment operator.
This means that the value of n is decremented by 1 before it is returned. Therefore, fun() returns 3, which is the original value of n before decrementing. The main function calls fun() twice and adds the results, which gives 3 + 3 = 6. Then, the main function prints the result with a %d format specifier, which shows the integer part of the result. Therefore, the output of the program is:
fun() = 3 fun() = 3 printf("%d \n", fun() + fun()) = 6 = 3
NEW QUESTION # 23
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 # 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. The program outputs 24
- B. The program outputs 4
- C. Compilation fails
- D. The program outputs 16
- E. The program outputs 8
Answer: B
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
What happens if you try to compile and run this program?
#include <stdio.h>
int main (int argc, char *argv[]) {
int main, Main, mAIN = 1;
Main = main = mAIN += 1;
printf ("%d", MaIn) ;
return 0;
}
Choose the right answer:
- A. The program outputs 3
- B. The program outputs 1
- C. The program outputs an unpredictable value
- D. Compilation fails
- E. The program outputs 2
Answer: D
Explanation:
The program is not a valid C program and cannot be compiled successfully. The reason is that the program uses the same name main for both a function and a variable, which is not allowed in C. The name main is a reserved keyword that denotes the entry point of the program, and it cannot be redefined or reused for any other purpose. Therefore, the compiler will report an error and the program will not run. References = C - main() function - Tutorialspoint, C Keywords - GeeksforGeeks, C Basic Syntax
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 / 1 + 4 / 2;
printf("%d",i);
return 0;
}
Choose the right answer:
- A. The program outputs 3
- B. The program outputs 4
- C. Compilation fails
- D. The program outputs 0
- E. The program outputs 5
Answer: B
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 # 27
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 a value less than 128
- B. Compilation fails
- C. The program outputs 128
- D. The program outputs a value greater than 128
- E. The program enters an infinite loop
Answer: A
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 # 28
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 1
- C. Compilation fails
- D. The program outputs 4
- E. The program outputs 5
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 # 29
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 three lines of text
- B. The program outputs "[]"
- C. The program outputs two lines of text
- D. The program outputs [John Bean]
- E. The program outputs nothing
Answer: D
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 # 30
What happens if you try to compile and run this program?
#include <stdio.h>
#include <string.h>
int main (int argc, char *argv[]) {
int a = 0, b = 1, c;
c = a++ && b++;
printf("%d",b);
return 0;
}
Choose the right answer:
- A. The program outputs 3
- B. Compilation fails
- C. The program outputs 2
- D. The program outputs 0
- E. The program outputs 1
Answer: E
Explanation:
he expression a++ && b++ involves the logical AND (&&) operator. In C, the logical AND op-erator short-circuits, meaning that if the left operand (a++ in this case) is false, the right operand (b++) is not evaluated.
Initially, a is 0, and b is 1. The result of a++ is 0 (false), so b++ is not evaluated. The value of b remains 1. The printf statement then prints the value of b, which is 1.
Therefore, the correct answer is "The program outputs 1."
References = CLA - C Associate Programmer documents
NEW QUESTION # 31
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 -1
- C. The program outputs 7
- D. The program outputs 0
- E. The program outputs 1
Answer: E
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 # 32
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