100% Pass Guaranteed Free CLA-11-03 Exam Dumps Apr 16, 2024
Verified & Latest CLA-11-03 Dump Q&As with Correct Answers
NEW QUESTION # 10
-
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 1
- B. The program outputs 0
- C. Compilation fails
- D. The program outputs 2
- E. The program outputs 3
Answer: A
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 # 11
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 three lines of text
- C. The program outputs nothing
- D. The program outputs "[]"
- E. The program outputs two lines of text
Answer: A
NEW QUESTION # 12
What happens if you try to compile and run this program?
#include <stdio.h>
int main (int argc, char *argv[]) {
char i = 20 + 020 + 0x20;
printf("%d",i);
return 0;
}
Choose the right answer:
- A. The program outputs 68
- B. The program outputs 60
- C. The program outputs 62
- D. Compilation fails
- E. The program outputs 86
Answer: A
Explanation:
*The program is a valid C program that can be compiled and run without errors.
*The variable i is declared as a char, which is an 8-bit signed integer type that can store val-ues from -128 to
127.
*The expression 20 + 020 + 0x20 evaluates to 68, because:
o20 is a decimal literal with the value 20
o020 is an octal literal with the value 16 (8^1 * 2 + 8^0 * 0)
o0x20 is a hexadecimal literal with the value 32 (16^1 * 2 + 16^0 * 0)
oThe + operator performs arithmetic addition on the operands and returns the sum
*The printf function prints the value of i as a decimal integer using the %d format specifier.
*The output of the program is 68.
NEW QUESTION # 13
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 10
- C. Compilation fails
- D. The program outputs 20
- 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 # 14
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. Compilation fails
- B. The program outputs 3.1415
- C. Execution fails
- D. The program outputs 3
- E. The program outputs 4
Answer: A
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 # 15
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 200
- B. The program outputs 400
- C. The program outputs 300
- D. The program outputs 100
- 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 # 16
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 1
- B. The program outputs 0
- C. Compilation fails
- D. The program outputs 2
- E. The program outputs 3
Answer: A
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 # 17
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 3
- B. Compilation fails
- C. The program outputs 2
- D. Execution fails
- E. The program outputs 4
Answer: A
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 # 18
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 1
- C. The program outputs 0
- D. Compilation fails
- E. The program outputs 7
Answer: B
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 # 19
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. Compilation fails
- B. The program outputs 12
- C. Execution fails
- D. The program outputs 4
- E. The program outputs 16
Answer: A
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 # 20
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. Compilation fails
- B. The program outputs -1
- C. The program outputs 1
- D. The program outputs -2
- E. The program outputs 2
Answer: A
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 # 21
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. The program outputs 0
- C. Compilation fails
- D. The program outputs 2
- E. The program executes an infinite loop
Answer: E
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 # 22
What happens if you try to compile and run this program?
enum { A, B, C, D, E, F };
#include <stdio.h>
int main (int argc, char *argv[]) {
printf ("%d", B + D + F);
return 0;
}
Choose the right answer:
- A. The program outputs 8
- B. The program outputs 10
- C. Compilation fails
- D. The progham outputs 9
- E. The program outputs 7
Answer: D
Explanation:
The program outputs 9 because the expression B + D + F evaluates to 9 using the enumeration constants defined by the enum keyword. The enum keyword creates a user-defined data type that can have one of a set of named values. By default, the first value is assigned 0, and each subsequent val-ue is assigned one more than the previous one, unless explicitly specified. Therefore, in this pro-gram, A is 0, B is 1, C is 2, D is 3, E is
4, and F is 5. The printf function then prints the sum of B, D, and F, which is 1 + 3 + 5 = 9, as a decimal integer using the %d format specifier.
References = CLA - C Certified Associate Programmer Certification, [C Essentials 2 - (Intermediate)], C Enumeration
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 / 1 + 4 / 2;
printf("%d",i);
return 0;
}
Choose the right answer:
- A. The program outputs 0
- B. The program outputs 5
- C. Compilation fails
- D. The program outputs 3
- 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 # 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/=2)
if(i)
break ;
printf("%d",i);
return 0;
}
Choose the right answer:
The program executes an infinite loop
- A. The program outputs 1
- B. The program outputs 0
- C. Compilation fails
- D. The program outputs 0.5
Answer: A
Explanation:
The program outputs 1 because the for loop terminates when i becomes 0. The for loop has no initialization, condition, or increment expressions, so it will run indefinitely unless a break statement is executed. The loop body consists of a single if statement that checks if i is non-zero,and if so, breaks out of the loop. Otherwise, i is divided by 2 and assigned back to itself. Since i is an integer, the division will truncate any fractional part.
Therefore, the loop will iterate until i becomes 0, which will happen after one iteration, as 1 / 2 = 0. The printf function then prints the value of i as a deci-mal integer using the %d format specifier.
References = CLA - C Certified Associate Programmer Certification, [C Essentials 2 - (Intermediate)], C For Loop, C If...Else Statement
NEW QUESTION # 25
What is the meaning of the following declaration?
float ** p;
Choose the right answer:
- A. p is a pointer to a pointer to a float
- B. p is a pointer to a float pointer
- C. p is a pointer to a float
- D. The declaration is erroneous
- E. p is a float pointer to a float
Answer: A
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 # 26
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