Trouble Shoot Round-1 Set-1

1. Int main() { int c=2; switch(c) { case 0: printf("CSE"); case 1: printf("Quest"); case 2: printf("Trouble"); case 3: printf("Shoot"); default: printf("2k19"); } return 0; }

CSEQuest2k19

TroubleShoot2k19

2k19

Quest2k19

The correct answer is "TroubleShoot2k19" because the switch statement is evaluating the value of the variable c. Since c is equal to 2, it matches the case 2 and prints "Trouble". However, there are no break statements after each case, so the execution continues to the next cases. As a result, it also prints "Shoot" and "2k19". Therefore, the final output is "TroubleShoot2k19".

Explanation

The correct answer is "TroubleShoot2k19" because the switch statement is evaluating the value of the variable c. Since c is equal to 2, it matches the case 2 and prints "Trouble". However, there are no break statements after each case, so the execution continues to the next cases. As a result, it also prints "Shoot" and "2k19". Therefore, the final output is "TroubleShoot2k19".

2. Int main() { int x=5; if(x>=10) printf("Quest"); printf("CSE"); else printf("2K19"); }

No Output

2K19

QuestCSE

Compilation Error

The given code will not produce a compilation error. The output will be "2K19". The if statement checks if the value of x is greater than or equal to 10, which is false in this case. Therefore, the code inside the if statement will not be executed. The code outside the if statement will be executed, which is the printf("2K19") statement, resulting in the output "2K19".

Explanation

The given code will not produce a compilation error. The output will be "2K19". The if statement checks if the value of x is greater than or equal to 10, which is false in this case. Therefore, the code inside the if statement will not be executed. The code outside the if statement will be executed, which is the printf("2K19") statement, resulting in the output "2K19".

3. How many times "Trouble Shoot" will print? int main() { int x; for(x=-1; x<=10; x++) { if(x < 5) continue; else break; printf("Trouble Shoot"); } return 0; }

Infinite Time

11 Times

0 Time

Error

The loop will run 6 times because x starts at -1 and increments by 1 each time until it reaches 5. However, when x is 5, the condition in the if statement is not satisfied, so the loop is exited and "Trouble Shoot" is not printed at all. Therefore, the correct answer is 0 times.

Explanation

The loop will run 6 times because x starts at -1 and increments by 1 each time until it reaches 5. However, when x is 5, the condition in the if statement is not satisfied, so the loop is exited and "Trouble Shoot" is not printed at all. Therefore, the correct answer is 0 times.

4. 3. int main() { char arr[5]="Quest"; printf("%s", arr); return 0; }

Quest

Q

Garbage Value

Compilation Error

The code declares an array of characters called "arr" with a size of 5. The string "Quest" is assigned to the array. The printf function is then used to print the string stored in "arr". Therefore, the output of the code will be "Quest".

Explanation

The code declares an array of characters called "arr" with a size of 5. The string "Quest" is assigned to the array. The printf function is then used to print the string stored in "arr". Therefore, the output of the code will be "Quest".

5. Int main(void) { while(.) { printf("Quest"); } return 0; }

Quest(Infinity Times)

Error

Segmentation Fault

None of the Above

The given code will result in an error because the condition inside the while loop is not specified. The while loop requires a condition to be evaluated, and since there is no condition provided (the dot symbol is not a valid condition), it will result in a syntax error.

Explanation

The given code will result in an error because the condition inside the while loop is not specified. The while loop requires a condition to be evaluated, and since there is no condition provided (the dot symbol is not a valid condition), it will result in a syntax error.

6. Int main() { int a = 100, b = 74; if (a++ > 100 && b++ > 200) printf("High values with a = %d b = %dn", a, b); if (a++ < 100 || b++ < 200) printf("Low values with a = %d b = %dn", a, b); }

Low values with a = 100 b = 74

Low values with a = 100 b = 73

Low values with a = 102 b = 75

Low values with a = 102 b = 74

The correct answer is "Low values with a = 102 b = 75" because the if statement `(a++

Explanation

The correct answer is "Low values with a = 102 b = 75" because the if statement `(a++

7. Void PrintCount() { static int Count=1; printf("%d ",Count); Count=Count+1; } int main() { PrintCount(); PrintCount(); PrintCount(); return 0; }

1 1 1

1 2 3

Compile time error

Run time error

The function PrintCount is called three times in the main function. The variable Count is declared as a static variable inside the PrintCount function, which means it retains its value between function calls.

Initially, Count is set to 1. The first time PrintCount is called, it prints the value of Count (which is 1) and then increments it by 1. So the first output is 1.

The second time PrintCount is called, it prints the value of Count (which is now 2) and then increments it by 1. So the second output is 2.

The third time PrintCount is called, it prints the value of Count (which is now 3) and then increments it by 1. So the third output is 3.

Therefore, the correct answer is "1 2 3".

Explanation

The function PrintCount is called three times in the main function. The variable Count is declared as a static variable inside the PrintCount function, which means it retains its value between function calls.

Initially, Count is set to 1. The first time PrintCount is called, it prints the value of Count (which is 1) and then increments it by 1. So the first output is 1.

The second time PrintCount is called, it prints the value of Count (which is now 2) and then increments it by 1. So the second output is 2.

The third time PrintCount is called, it prints the value of Count (which is now 3) and then increments it by 1. So the third output is 3.

Therefore, the correct answer is "1 2 3".

8. Void test(struct number n) { n.x=100; } struct number{ int x; }; int main() { struct number num; test(num); printf("%d\n",num.x); return 0; }

Error

No Output

100

None of the Above

The code is giving an error because the function `test` is trying to modify the value of `num.x`, but it is passed by value to the function. This means that the function is working with a copy of `num`, and any changes made to `n.x` inside the function will not affect the original `num.x`. To fix this error, the function `test` should take a pointer to `struct number` as a parameter, allowing it to modify the original `num.x`.

Explanation

The code is giving an error because the function `test` is trying to modify the value of `num.x`, but it is passed by value to the function. This means that the function is working with a copy of `num`, and any changes made to `n.x` inside the function will not affect the original `num.x`. To fix this error, the function `test` should take a pointer to `struct number` as a parameter, allowing it to modify the original `num.x`.

9. #include<stdio.h> #include<stdlib.h> int main() { unsigned char; FILE *fp; fp=fopen("trial", "r"); if(!fp) { printf("Unable to open file"); exit(1); } fclose(fp); return 0; }

Error: in unsigned char statement

Error: unknown file pointer

No Error

None of above

The given code does not contain any syntax errors or logical errors. The code opens a file named "trial" in read mode using the fopen() function. If the file does not exist or cannot be opened, the code prints "Unable to open file" and exits the program. Otherwise, the code closes the file using the fclose() function and returns 0. Therefore, there is no error in the code.

Explanation

The given code does not contain any syntax errors or logical errors. The code opens a file named "trial" in read mode using the fopen() function. If the file does not exist or cannot be opened, the code prints "Unable to open file" and exits the program. Otherwise, the code closes the file using the fclose() function and returns 0. Therefore, there is no error in the code.

10. Int main() { int x[10]={0,1,2,3,4,5,6,7,8,9}; int *ptr1,*ptr2; ptr1=&x[0]; ptr2=&x[5]; printf("%p\n",(ptr1+ptr2)); return 0; }

Compilation Error

Garbage Value

Some Random Address

No Error

The expression `(ptr1+ptr2)` is not valid because it is trying to add two pointers together. Pointer arithmetic can only be performed when adding or subtracting an integer value to a pointer. In this case, `ptr1` and `ptr2` are both pointers to integers, so adding them together is not allowed. Therefore, the code will not compile and a compilation error will occur.

Explanation

The expression `(ptr1+ptr2)` is not valid because it is trying to add two pointers together. Pointer arithmetic can only be performed when adding or subtracting an integer value to a pointer. In this case, `ptr1` and `ptr2` are both pointers to integers, so adding them together is not allowed. Therefore, the code will not compile and a compilation error will occur.

11. Int main() { int x = 0, y = 0; if(x > 0) if(y > 0) printf("True"); else printf("False"); }

No Output

True

False

Error Because of dangling else problem

The correct answer is "No Output" because the condition `x > 0` is not satisfied, so the inner `if` statement is not executed. Therefore, there is no output printed.

Explanation

The correct answer is "No Output" because the condition `x > 0` is not satisfied, so the inner `if` statement is not executed. Therefore, there is no output printed.

12. Int main() { int x,y,z; x = '1'-'0'; /* line-1 */ y = 'a'-'b'; /* line-2 */ z = x + y; printf("%d",z); }

0

Error because of incorrect line-1 only

Error because of incorrect line-1 and line-2.

Error because of incorrect line-2 only.

The correct answer is 0 because the expression '1'-'0' in line-1 will result in the integer value 1-0=1. Similarly, the expression 'a'-'b' in line-2 will result in the integer value 97-98=-1. Therefore, the value of x is 1 and the value of y is -1. When x and y are added together in line-3, the result is 1+(-1)=0. Finally, the value of z is printed, which is 0.

Explanation

The correct answer is 0 because the expression '1'-'0' in line-1 will result in the integer value 1-0=1. Similarly, the expression 'a'-'b' in line-2 will result in the integer value 97-98=-1. Therefore, the value of x is 1 and the value of y is -1. When x and y are added together in line-3, the result is 1+(-1)=0. Finally, the value of z is printed, which is 0.

13. #include<stdio.h> int main() { int i; for(i=0;i<=3;i++); printf("%d",i); return 0; }

Compilation Error

1 2 3

4

0 1 2 3

The correct answer is 4.

In this code, there is a for loop that runs from i=0 to i
After the loop, the printf statement is executed, which prints the value of i, which is 4 because the loop did not change the value of i. Therefore, the output of the code is 4.

Explanation

The correct answer is 4.

In this code, there is a for loop that runs from i=0 to i
After the loop, the printf statement is executed, which prints the value of i, which is 4 because the loop did not change the value of i. Therefore, the output of the code is 4.

14. Int main() { char *str="Quest"; printf("%d,%d\n",strlen(str),sizeof(str)); return 0; }

5,10

5,8

5,6

5,5

The code snippet declares a character pointer variable `str` and assigns it the address of the string "Quest". The `strlen()` function is then used to calculate the length of the string, which is 5. The `sizeof()` function is used to determine the size of the pointer variable `str`, which is 8 bytes on most systems. Therefore, the output of the `printf()` function will be "5,8".

Explanation

The code snippet declares a character pointer variable `str` and assigns it the address of the string "Quest". The `strlen()` function is then used to calculate the length of the string, which is 5. The `sizeof()` function is used to determine the size of the pointer variable `str`, which is 8 bytes on most systems. Therefore, the output of the `printf()` function will be "5,8".

15. Void main() { int a,b,c; a+b=c; } What type of error we get

Syntax Error

Semantic Error

Logical Error

Run Time Error

The given code has a semantic error. In the line "a+b=c;", the assignment operator (=) is used incorrectly. The left side of the assignment operator should be a variable, but in this case, it is an expression (a+b). This violates the syntax rules of the programming language and results in a semantic error.

Explanation

The given code has a semantic error. In the line "a+b=c;", the assignment operator (=) is used incorrectly. The left side of the assignment operator should be a variable, but in this case, it is an expression (a+b). This violates the syntax rules of the programming language and results in a semantic error.

16. Int main() { int a; int b=5; a=0 && --b; printf("%d%d",a,b); }

0 4

Compile Time Error

0 5

Run Time Error

In this code, the variable "a" is assigned the result of the logical AND operation between 0 and the pre-decremented value of "b". Since the value of "b" is initially 5, it is decremented to 4 before the logical AND operation is performed. The logical AND operator returns 0 if any of the operands is 0, otherwise it returns 1. In this case, the result of the logical AND operation is 0 because one of the operands is 0. Therefore, "a" is assigned the value 0. The value of "b" remains 5 because the decrement operation is not assigned to it. Finally, the printf statement prints the values of "a" and "b", which are 0 and 5 respectively.

Explanation

In this code, the variable "a" is assigned the result of the logical AND operation between 0 and the pre-decremented value of "b". Since the value of "b" is initially 5, it is decremented to 4 before the logical AND operation is performed. The logical AND operator returns 0 if any of the operands is 0, otherwise it returns 1. In this case, the result of the logical AND operation is 0 because one of the operands is 0. Therefore, "a" is assigned the value 0. The value of "b" remains 5 because the decrement operation is not assigned to it. Finally, the printf statement prints the values of "a" and "b", which are 0 and 5 respectively.

17. #include<stdio.h> int main() { char ch; int i; scanf("%c", &i); scanf("%d", &ch); printf("%c %d", ch, i); return 0; }

Error: suspicious char to in conversion in scanf()

Error: we may not get input for second scanf() statement

No error

None of above

The correct answer is "Error: we may not get input for second scanf() statement". This is because the format specifier used in the first scanf() statement is incorrect. The variable 'i' is an integer, but the format specifier used is '%c' which is for characters. This will cause an error and the input for 'i' will not be read correctly. As a result, the second scanf() statement may not receive any input, leading to an error.

Explanation

The correct answer is "Error: we may not get input for second scanf() statement". This is because the format specifier used in the first scanf() statement is incorrect. The variable 'i' is an integer, but the format specifier used is '%c' which is for characters. This will cause an error and the input for 'i' will not be read correctly. As a result, the second scanf() statement may not receive any input, leading to an error.

18. #include<stdio.h> #include<stdlib.h> int main() { char *ptr; free(ptr); return 0 }

This program will print nothing after execution

Segmentation fault

Aborted (core dumped)

None of the mentioned

The program is attempting to free a pointer variable without allocating any memory to it. This results in undefined behavior and can lead to a segmentation fault or an aborted program with a core dump. In this case, the program is likely to be aborted and a core dump is generated.

Explanation

The program is attempting to free a pointer variable without allocating any memory to it. This results in undefined behavior and can lead to a segmentation fault or an aborted program with a core dump. In this case, the program is likely to be aborted and a core dump is generated.

19. Void Main() { int a=10; printf("%d", a); }

10

Run Time Error

Compilation Error

None of the Above

The given code is written in C language. It initializes a variable 'a' with the value 10 and then prints the value of 'a' using the printf function. Since the code does not have any syntax errors, it will compile successfully. However, it will result in a runtime error because the format specifier used in the printf function is incorrect. The correct format specifier for an integer is "%d" but in the code, it is written as "%quot;%d%quot;". This will cause the code to throw a runtime error.

Explanation

The given code is written in C language. It initializes a variable 'a' with the value 10 and then prints the value of 'a' using the printf function. Since the code does not have any syntax errors, it will compile successfully. However, it will result in a runtime error because the format specifier used in the printf function is incorrect. The correct format specifier for an integer is "%d" but in the code, it is written as "%quot;%d%quot;". This will cause the code to throw a runtime error.

20. Examination of the program step by step is called ______________

Controlling

Stepping

Tracing

Testing

Stepping refers to the process of examining a program step by step, allowing the programmer to analyze the execution flow and identify any errors or bugs. It involves executing the program line by line, pausing at each step to observe the variables, data values, and program state. This method helps in understanding the program's behavior and identifying the exact point where an error occurs. By stepping through the code, programmers can effectively debug and troubleshoot their programs.

Explanation

Stepping refers to the process of examining a program step by step, allowing the programmer to analyze the execution flow and identify any errors or bugs. It involves executing the program line by line, pausing at each step to observe the variables, data values, and program state. This method helps in understanding the program's behavior and identifying the exact point where an error occurs. By stepping through the code, programmers can effectively debug and troubleshoot their programs.