C++ Institute CLA-11-03 Real Exam Questions Guaranteed Updated Dump from Actualtests4sure
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NEW QUESTION # 15
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 two lines of text
- C. The program outputs [John Bean]
- D. The program outputs "[]"
- E. The program outputs nothing
Answer: C
NEW QUESTION # 16
Assume that we can open a file called "file1".
What happens when you try to compile and run the following program?
#include <stdio.h>
int main (void) {
FILE *f;
int i;
f = fopen("file1","wb");
fputs("545454",f);
fclose (f);
f = fopen("file1","rt");
fscanf(f,"%d ", &i);
fclose (f) ;
printf("%d",i);
return 0;
}
Choose the right answer:
- A. The program outputs 0
- B. The program outputs 54
- C. Execution fails
- D. Compilation fails
- E. The program outputs 545454
Answer: E
Explanation:
The program outputs 545454 because the fputs function writes the string "545454" to the file "file1" in binary mode, and the fscanf function reads the string as an integer from the file in text mode. The binary mode and the text mode are different ways of interpreting the data in a file. In binary mode, the data is stored as a sequence of bytes, and no translation is performed. In text mode, the data is stored as a sequence of characters, and some characters may be translated depending on the plat-form. For example, the newline character may be translated to a carriage return and a line feed on Windows, or just a line feed on Linux. The fopen function takes a mode argument that specifies whether the file should be opened in binary or text mode. The mode
"wb" means write binary, and the mode "rt" means read text.
When the fputs function writes the string "545454" to the file in binary mode, it writes the ASCII val-ues of each character as a byte. The ASCII value of '5' is 53, and the ASCII value of '4' is 52. There-fore, the file contains the following bytes: 53 53 53 52 52 52. When the fscanf function reads the file in text mode, it interprets the bytes as characters and converts them to an integer using the %d format specifier. The %d format specifier expects a decimal integer, which is a number com-posed of digits from 0 to 9. Since the file contains only digits, the fscanf function successfully con-verts the string "545454" to an integer with the same value.
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 File I/O, ASCII Table
NEW QUESTION # 17
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. The program outputs 0
- C. Compilation fails
- D. The program outputs 1
- E. The program outputs 2
Answer: D
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 # 18
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 8
- B. The program outputs 16
- C. Compilation fails
- D. The program outputs 24
- 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 # 19
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 0
- C. The program outputs 5
- D. Compilation fails
- E. The program outputs 4
Answer: E
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 # 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 0
- C. The program outputs 9
- D. Execution fails
- E. Compilation fails
Answer: B
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[]) {
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 # 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 three lines of text
- B. The program outputs two lines of text
- C. The program outputs [John Bean]
- D. The program outputs "[]"
- E. The program outputs nothing
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[]) {
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. The program outputs 6
- C. The program outputs 4
- D. Compilation fails
- 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 # 24
What is the meaning of the following declaration?
float ** p;
Choose the right answer:
- A. p is a pointer to a float
- B. p is a float pointer to a float
- C. p is a pointer to a pointer to a float
- D. The declaration is erroneous
- E. p is a pointer to a float pointer
Answer: C
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
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 14
- B. The program outputs 20
- C. The program outputs 10
- D. Compilation fails
- E. The program outputs 24
Answer: A
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 # 26
What happens if you try to compile and run this program?
#define ALPHA 0
#define BETA ALPHA-1
#define GAMMA 1
#define dELTA ALPHA-BETA-GAMMA
#include <stdio.h>
int main(int argc, char *argv[]) {
printf ("%d", DELTA);
return 0;
Choose the right answer:
- A. The program outputs 1
- B. The program outputs -1
- C. Compilation fails
- D. The program outputs -2
- E. The program outputs 2
Answer: C
Explanation:
Let's analyze the macros and the program:
1.ALPHA is defined as 0.
2.BETA is defined as ALPHA - 1, which is 0 - 1.
3.GAMMA is defined as 1.
4.DELTA is defined as ALPHA - BETA - GAMMA. With the previous definitions, this expands to 0 - (0 - 1) -
1.
Now, let's expand DELTA with the given values:
makefileCopy code
DELTA = 0 - (0 - 1) - 1 DELTA = 0 - 0 + 1 - 1 DELTA = 0 + 1 - 1 DELTA = 1 - 1 DELTA = 0 It is important to note that the macro dELTA is defined with a lowercase 'd', but the printf function is trying to print DELTA with an uppercase 'D'. Preprocessor tokens are case-sensitive, so this is a mismatch. However, for the sake of the question, let's assume that dELTA was meant to be DELTA with an uppercase 'D'.
Since the actual calculation results in 0, but there is a typo in the printf statement (it should print dELTA, not DELTA), the compilation will fail due to DELTA not being defined.
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 greater than 128
- B. The program outputs a value less than 128
- C. The program outputs 128
- D. Compilation fails
- E. The program enters an infinite loop
Answer: B
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
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 # 29
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 1
- C. Compilation fails
- D. The program outputs 4
- E. The program outputs 2
Answer: D
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 # 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 3
- B. The program outputs 3.1415
- C. Execution fails
- 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 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
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