Should we break the default case in switch statement?

Question

Assuming this example code (source):

#include <stdio.h>

void playgame()
{
    printf( "Play game called" );
}
void loadgame()
{
    printf( "Load game called" );
}
void playmultiplayer()
{
    printf( "Play multiplayer game called" );
}

int main()
{
    int input;

    printf( "1. Play game\n" );
    printf( "2. Load game\n" );
    printf( "3. Play multiplayer\n" );
    printf( "4. Exit\n" );
    printf( "Selection: " );
    scanf( "%d", &input );
    switch ( input ) {
        case 1:            /* Note the colon, not a semicolon */
            playgame();
            break;
        case 2:
            loadgame();
            break;
        case 3:
            playmultiplayer();
            break;
        case 4:
            printf( "Thanks for playing!\n" );
            break;
        default:
            printf( "Bad input, quitting!\n" );
            break;
    }
    getchar();

    return 0;
}

should we use a break; in the last default case? If I remove it, I see the same behaviour of the program. However, I saw that other examples also use a break; in the default case.

Why? Is there a reason?

Answer 1

Should we use a break; in the last default case?

From The C programming language - Second edition (K&R 2):

Chapter 3.4 Switch

As a matter of good form, put a break after the last case (the default here) even though it's logically unnecessary. Some day when another case gets added at the end, this bit of defensive programming will save you.