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Pass CLA-11-03 Exam with 41 Questions - Verified By TorrentValid
NEW QUESTION # 11
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 1
- B. The program outputs 2
- C. The program outputs an unpredictable value
- D. Compilation fails
- E. The program outputs 0
Answer: A
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 # 12
Select the proper form for the following declaration:
p is a pointer to an array containing 10 int values
Choose the right answer:
- A. The declaration is invalid and cannot be coded in C
- B. int *p[10];
- C. int * (p) [10];
- D. int (*p) [10];
- E. int (*)p[10];
Answer: D
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 # 13
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 54
- B. Execution fails
- C. Compilation fails
- D. The program outputs 545454
- E. The program outputs 0
Answer: D
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 # 14
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 pointer to a float pointer
- C. p is a float pointer to a float
- D. The declaration is erroneous
- E. p is a pointer to a pointer to a float
Answer: E
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 # 15
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 4
- B. The program outputs 3
- C. The program outputs 5
- D. Compilation fails
- E. The program outputs 0
Answer: A
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 # 16
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 4
- B. The program outputs 3
- C. The program outputs 2
- D. The program outputs 1
- E. The program outputs 0
Answer: C
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 # 17
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. Compilation fails
- D. The program outputs -1
- E. The program outputs 1
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 # 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. Compilation fails
- D. The program outputs 0
- E. The program outputs 7
Answer: A
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
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 progham outputs 9
- B. The program outputs 8
- C. Compilation fails
- D. The program outputs 10
- E. The program outputs 7
Answer: A
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 # 20
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 4
- B. The program outputs 9
- C. The program outputs 15
- D. Compilation fails
- E. The program outputs 0
Answer: B
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 # 21
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 100
- B. The program outputs 600
- C. The program outputs 200
- D. The program outputs 300
- E. The program outputs 400
Answer: B
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 # 22
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 2
- C. The program outputs 3
- D. Compilation fails
- E. The program outputs 0
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 # 23
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 executes an infinite loop
- B. The program outputs 2
- C. Compilation fails
- D. The program outputs 1
- E. The program outputs 0
Answer: A
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 # 24
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 9
- B. The program outputs 3
- C. Execution fails
- D. Compilation fails
- E. The program outputs 0
Answer: E
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 # 25
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 12300.000
- B. The program outputs 12.300000
- C. The program outputs 1230.0000
- D. Compilation fails
- E. The program outputs 123.00000
Answer: B
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 # 26
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 2
- B. Compilation fails
- C. The program outputs 3
- D. The program outputs 1
- E. The program outputs 0
Answer: B
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 # 27
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