Latest [Apr 18, 2024] CLA-11-03 Exam Dumps - Valid and Updated Dumps
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NEW QUESTION # 12
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. Compilation fails
- C. The program outputs 5
- D. The program outputs 1
- E. The program outputs 4
Answer: E
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 # 13
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 3
- C. The program outputs 2
- D. The program outputs 4
- E. The program outputs 1
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 # 14
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. Compilation fails
- B. The program outputs 10
- C. The program outputs 20
- D. The program outputs 14
- E. The program outputs 24
Answer: D
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 # 15
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 12
- B. Execution fails
- C. The program outputs 16
- 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 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 # 16
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 -2
- B. The program outputs 2
- C. The program outputs -1
- D. The program outputs 1
- E. Compilation fails
Answer: E
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 # 17
What happens if you try to compile and run this program?
#include <stdio.h>
#include <stdlib.h>
int main (int argc, char *argv[]) {
double x = 1234567890.0;
printf ("%f",x);
return 0;
}
Choose the most precise answer:
- A. The program outputs 1234567900.0
- B. Compilation fails
- C. The program outputs a value greater than 1234500000.0 and less than 1234600000.0
- D. Execution fails
- E. The program outputs 1234567890.0
Answer: E
Explanation:
To understand the behavior of this program, let's first analyze its structure:
1.It includes the standard I/O library, <stdio.h>, and the standard library, <stdlib.h>, although <stdlib.h> is not used in this program.
2.main function declares a double variable x initialized to 1234567890.0.
3.It then prints x using %f format specifier in printf.
Key Points to Consider:
*The double data type in C is typically capable of representing a wide range of deci-mal numbers quite accurately.
*The %f format specifier in printf is used for outputting a float or double as a fixed-point number.
*There may be some precision issues when dealing with floating-point numbers, but these generally occur with more complex calculations or when the numbers are ex-tremely large or small.
Given that 1234567890.0 is a straightforward decimal number well within the representable range of a double, and the program doesn't perform any complex calculations, we can ex-pect the output to be quite close to the actual value of x.
However, due to the way floating-point numbers are represented and handled in C, there can be slight discrepancies in the least significant digits due to rounding or representation errors.
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: 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 # 19
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 nothing
- C. The program outputs two lines of text
- D. The program outputs "[]"
- E. The program outputs three lines of text
Answer: A
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. Compilation fails
- B. The program outputs 123.00000
- C. The program outputs 1230.0000
- D. The program outputs 12300.000
- E. The program outputs 12.300000
Answer: E
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
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. Compilation fails
- B. The program outputs 128
- C. The program outputs a value greater than 128
- D. The program outputs a value less than 128
- E. The program enters an infinite loop
Answer: D
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 # 22
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. The program outputs 3
- C. Execution fails
- D. The program outputs 2
- E. The program outputs 4
Answer: B
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 # 23
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 3
- B. The program outputs 5
- C. The program outputs 1
- D. The program outputs 2
- E. The program outputs 4
Answer: C
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 # 24
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. Compilation fails
- C. The program outputs 0
- D. The program outputs 5
- 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 # 25
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. The program outputs 0
- C. The program outputs -1
- D. Execution fails
- E. The program outputs 1
Answer: C
NEW QUESTION # 26
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. Compilation fails
- B. The program outputs []
- C. The program outputs []
- D. Execution fails
- E. The program outputs ["]
Answer: D
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 # 27
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 0
- C. The program outputs an unpredictable value
- D. The program outputs 1
- E. The program outputs 2
Answer: D
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 # 28
......
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