Coding And Debugging - Round 1- Set B

1. #include void main() { int i = 0; while (++i) { printf("H"); } }

H

H is printed infinte times

Compile time error

Varies

The given code is an infinite loop because the condition of the while loop is always true. The variable "i" is incremented by 1 before each iteration of the loop. Since the value of "i" is initially 0, it becomes 1 in the first iteration, which satisfies the condition of the while loop. Therefore, the statement "H" is printed infinitely.

Explanation

The given code is an infinite loop because the condition of the while loop is always true. The variable "i" is incremented by 1 before each iteration of the loop. Since the value of "i" is initially 0, it becomes 1 in the first iteration, which satisfies the condition of the while loop. Therefore, the statement "H" is printed infinitely.

2. #include int main() { int i = 0, j = 0; l1: while (i < 2) { i++; while (j < 3) { printf("loop\n"); goto l1; } } }

Loop loop

Compilation error

Loop loop loop loop

Infinte loop

The code starts with initializing the variables i and j to 0. Then it enters a while loop with the condition i

Explanation

The code starts with initializing the variables i and j to 0. Then it enters a while loop with the condition i

3. #include int main() { int a = 0, i = 0, b; for (i = 0;i < 5; i++) { a++; continue; printf("%d",a); } printf("%d",a); }

5

10

7

1

In this code, a variable 'a' is initialized to 0. Then, a for loop is executed 5 times. In each iteration, the value of 'a' is incremented by 1 using the 'a++' statement. The 'continue' statement is used to skip the remaining code inside the loop and move to the next iteration. Therefore, the printf statement inside the loop is never executed. After the loop, the final value of 'a' is printed, which is 5.

Explanation

In this code, a variable 'a' is initialized to 0. Then, a for loop is executed 5 times. In each iteration, the value of 'a' is incremented by 1 using the 'a++' statement. The 'continue' statement is used to skip the remaining code inside the loop and move to the next iteration. Therefore, the printf statement inside the loop is never executed. After the loop, the final value of 'a' is printed, which is 5.

4. #include void main() { int x = 0; if (x == 0) printf("hi"); else printf("how are u"); printf("hello"); }

Hi

How are you

Hello

Hihello

The given code snippet is a C program that prints "hi", "how are u", and "hello" on separate lines. The variable x is initialized to 0 and the if statement checks if x is equal to 0. Since x is indeed equal to 0, the code inside the if block is executed, which prints "hi". The code outside the if block, which is the printf statement for "hello", is always executed regardless of the if condition. Therefore, the output of the program is "hi" followed by "hello".

Explanation

The given code snippet is a C program that prints "hi", "how are u", and "hello" on separate lines. The variable x is initialized to 0 and the if statement checks if x is equal to 0. Since x is indeed equal to 0, the code inside the if block is executed, which prints "hi". The code outside the if block, which is the printf statement for "hello", is always executed regardless of the if condition. Therefore, the output of the program is "hi" followed by "hello".

5. #include void main() { int i = 0, j = 0; for (i = 0;i < 5; i++) { for (j = 0;j < 4; j++) { if (i > 1) break; } printf("Hi \n"); } }

Hi is printed 5 times

Hi is printed 9 times

Hi is printed 7 times

Hi is printed 4 times

The inner loop is never executed because the condition "i > 1" is never true. Therefore, the inner loop doesn't affect the number of times "Hi" is printed. The outer loop runs 5 times, so "Hi" is printed 5 times.

Explanation

The inner loop is never executed because the condition "i > 1" is never true. Therefore, the inner loop doesn't affect the number of times "Hi" is printed. The outer loop runs 5 times, so "Hi" is printed 5 times.

6. #include void main() { int i = 0; int j = 0; for (i = 0;i < 5; i++) { for (j = 0;j < 4; j++) { if (i > 1) continue; printf("Hi \n"); } } }

Hi is printed 9 times

Hi is printed 8 times

Hi is printed 7 times

Hi is printed 6 times

The code contains nested for loops. The outer loop runs 5 times and the inner loop runs 4 times. However, there is a conditional statement inside the inner loop that uses the "continue" keyword. This means that if the value of "i" is greater than 1, the inner loop will skip the rest of its iterations and move to the next iteration of the outer loop. Therefore, when "i" is 2, 3, or 4, the inner loop is skipped entirely. As a result, "Hi" is only printed during the first two iterations of the outer loop, which gives a total of 8 times.

Explanation

The code contains nested for loops. The outer loop runs 5 times and the inner loop runs 4 times. However, there is a conditional statement inside the inner loop that uses the "continue" keyword. This means that if the value of "i" is greater than 1, the inner loop will skip the rest of its iterations and move to the next iteration of the outer loop. Therefore, when "i" is 2, 3, or 4, the inner loop is skipped entirely. As a result, "Hi" is only printed during the first two iterations of the outer loop, which gives a total of 8 times.

7. #include int main(){int x = 10000;double y = 56;int *p = &x;double *q = &y;printf("p and q are %d and %d", sizeof(p), sizeof(q));return 0;}

P and q are 4 and 4

P and q are 4 and 8

Compiler error

P and q are 2 and 8

The correct answer is "p and q are 4 and 4". The sizeof() function in C returns the size in bytes of its operand. In this case, p is a pointer to an integer and q is a pointer to a double. Both of these pointer types have a size of 4 bytes on most systems. Therefore, the sizeof(p) and sizeof(q) both evaluate to 4.

Explanation

The correct answer is "p and q are 4 and 4". The sizeof() function in C returns the size in bytes of its operand. In this case, p is a pointer to an integer and q is a pointer to a double. Both of these pointer types have a size of 4 bytes on most systems. Therefore, the sizeof(p) and sizeof(q) both evaluate to 4.

8. #include int main() { float x = 'a'; printf("%f", x); return 0; } #include int main() { float x = 'a'; printf("%f", x); return 0; }

A

Run time error

A.0000000

97.000000

The correct answer is 97.000000 because in the code, the variable "x" is assigned the value of 'a', which is the ASCII value of the character 'a'. Since "x" is a float data type, it will be implicitly converted to its corresponding ASCII value, which is 97. When the printf function is called with the format specifier "%f", it will print the value of "x" as a floating-point number, which is 97.000000.

Explanation

The correct answer is 97.000000 because in the code, the variable "x" is assigned the value of 'a', which is the ASCII value of the character 'a'. Since "x" is a float data type, it will be implicitly converted to its corresponding ASCII value, which is 97. When the printf function is called with the format specifier "%f", it will print the value of "x" as a floating-point number, which is 97.000000.

9. #include int main() { float f1 = 0.1; if (f1 == 0.1) printf("equal\n"); else printf("not equal\n"); }

Equal

Not equal

Depend on compiler

None of the above

The reason why the answer is "not equal" is because floating-point numbers cannot be represented exactly in binary. The value 0.1 cannot be represented exactly in binary, so when it is assigned to the variable f1, it may not have the exact value of 0.1. Therefore, when comparing f1 to 0.1, they may not be equal.

Explanation

The reason why the answer is "not equal" is because floating-point numbers cannot be represented exactly in binary. The value 0.1 cannot be represented exactly in binary, so when it is assigned to the variable f1, it may not have the exact value of 0.1. Therefore, when comparing f1 to 0.1, they may not be equal.

10. #include void main() { int ch; printf("enter a value btw 1 to 2:"); scanf("%d", &ch); switch (ch) { case 1: printf("1\n"); break; printf("Hi"); default: printf("2\n"); } }

1

Hi 2

Run time error

2

The given code snippet is using a switch statement to check the value of the variable "ch". If the value is 1, it will print "1". However, after the break statement, there is a printf statement that will print "Hi". This statement is unreachable because the break statement will cause the control to exit the switch statement. Therefore, the code will only print "1".

Explanation

The given code snippet is using a switch statement to check the value of the variable "ch". If the value is 1, it will print "1". However, after the break statement, there is a printf statement that will print "Hi". This statement is unreachable because the break statement will cause the control to exit the switch statement. Therefore, the code will only print "1".