Latest [Jul 09, 2024] Realistic Verified CLA-11-03 Dumps
Pass C++ Institute CLA-11-03 Exam Updated 41 Questions
NEW QUESTION # 20
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 # 21
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. Compilation fails
- B. Execution fails
- C. The program outputs 0
- D. The program outputs 9
- E. The program outputs 3
Answer: C
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 # 22
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 # 23
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. Compilation fails
- B. Execution fails
- C. The program outputs 545454
- D. The program outputs 0
- E. The program outputs 54
Answer: C
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 # 24
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
NEW QUESTION # 25
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 # 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. 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 # 27
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 # 28
What happens if you try to compile and run this program?
#include <stdio.h>
int main (int argc, char *argv[]) {
int i = 'A' - 'B';
int j = 'b' - 'a';
printf("%d",i / j);
return 0;
}
Choose the right answer:
- A. Compilation fails
- B. Execution fails
- C. The program outputs -1
- D. The program outputs 0
- E. The program outputs 1
Answer: C
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 = 1, j = 0;
int 1 = !i + !! j;
printf("%d", 1);
return 0;
}
Choose the right answer:
- A. Compilation fails
- B. The program outputs 0
- C. The program outputs 3
- D. The program outputs 2
- E. The program outputs 1
Answer: A
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 # 30
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 0
- B. The program outputs 4
- C. The program outputs 3
- D. The program outputs 2
- E. The program outputs 1
Answer: C
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 # 31
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 # 32
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 # 33
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. Compilation fails
- B. The program outputs 5
- C. The program outputs 4
- D. The program outputs 3
- E. The program outputs 6
Answer: B
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 # 34
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 4
- C. The program outputs 3
- D. The program outputs 2
- E. The program outputs 1
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 # 35
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 # 36
......
Get 2024 Updated Free C++ Institute CLA-11-03 Exam Questions and Answer: https://www.pass4suresvce.com/CLA-11-03-pass4sure-vce-dumps.html