Question

What, if anything, is wrong with following code?

void test(int x)
{
switch (x)
{
case 1:
case 2:
case 0:
default:
case 4:
}
}
(a) The variable x does not have the right type for a switch expression
(b) The case label 0 must precede case label 1
(c) Each case section must end with a break statement
(d) The default label must be the last label in the switch statement
(e) There is nothing wrong with the code.

Answer 1

The default label must be the last label in the switch statement.

Step-by-step explanation:

The correct option is (d) The default label must be the last label in the switch statement.

In a switch statement, the default label (if present) should always be the last label. This is because the default label is executed if none of the other case labels match the value of the switch expression. If the default label is placed in the middle or anywhere else in the switch statement, it will cause a compilation error.

For example, in the given code, the default label is placed after the case label 4, which is incorrect. To fix the code, the default label should be the last label in the switch statement.

Answer 2

The default label must be the last label in the switch statement, s wrong with following code

Therefore, correct answer is (d) The default label must be the last label in the switch statement.

Step-by-step explanation:

The issue with the given code lies in the placement of the default label. According to C/C++ syntax rules for a switch statement, the default label must be the last label in the switch block. In the provided code, the default label appears before the case label 4, which is not allowed.

The corrected version of the code should have the default label placed at the end, after all the case labels, including case 4.

Corrected code snippet:

void test(int x)

{

switch (x)

{

case 1:

case 2:

case 0:

case 4:

break;

default:

break;

}

}

Therefore, correct answer is (d) The default label must be the last label in the switch statement.