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NEW QUESTION # 12
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 # 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. Compilation fails
- B. The program outputs 0
- C. Execution fails
- D. The program outputs 54
- 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 # 14
What happens if you try to compile and run this program?
#include <stdio.h>
int main (int argc, char *argv[]) {
int main, Main, mAIN = 1;
Main = main = mAIN += 1;
printf ("%d", MaIn) ;
return 0;
}
Choose the right answer:
- A. The program outputs 3
- B. The program outputs an unpredictable value
- C. The program outputs 2
- D. The program outputs 1
- 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 uses the same name main for both a function and a variable, which is not allowed in C. The name main is a reserved keyword that denotes the entry point of the program, and it cannot be redefined or reused for any other purpose. Therefore, the compiler will report an error and the program will not run. References = C - main() function - Tutorialspoint, C Keywords - GeeksforGeeks, C Basic Syntax
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 [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
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 # 16
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. Compilation fails
- B. The program outputs 0
- C. The program outputs 1
- D. The program outputs 0.5
Answer: C
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 # 17
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. Compilation fails
- C. The program outputs 6
- D. The program outputs 4
- 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 # 18
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. Compilation fails
- B. The program outputs 0
- C. The program outputs 2
- D. The program outputs 1
- E. The program outputs 4
Answer: E
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 # 19
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 3
- B. The program outputs 0
- C. The program outputs 2
- D. The program outputs 4
- E. The program outputs 1
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 # 20
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 # 21
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 # 22
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 400
- B. The program outputs 300
- C. The program outputs 100
- D. The program outputs 600
- E. The program outputs 200
Answer: D
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 # 23
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. Compilation fails
- C. The program outputs 0
- 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 # 24
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 # 25
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 3
- B. The program outputs 0
- C. The program outputs 2
- D. The program outputs 1
- E. Compilation fails
Answer: E
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 # 26
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 # 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. 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 # 28
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