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.

Answer 2

For one thing, you should think about why we should use break in switch statement. Look at this no-breaking switch statement.

switch ( input ) {     case 1:            /* Note the colon, not a semicolon */         playgame();     case 2:         loadgame();     case 3:         playmultiplayer();     case 4:         printf( "Thanks for playing!\n" );     default:         printf( "Bad input, quitting!\n" ); } 

Suppose input == 1. The program will call playgame() of course, but since there's no break, program won't finish the switch but call loadgame(), playmultiplayer(), two printfs sequentially.

To avoid this, we use break.

case 1:     playgame();     break; /* here */ case 2:     ... 

Because of break, the program finishes switch statement before running codes of case 2. That's our expected result, isn't it?

Your switch is this:

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; } 

Since there's no cases after default, there's no effect whether you write break on default or not. However, you can easily suppose to write a new case.

    default:         printf( "Thanks for playing!\n" );         /* what happens if there's no `break`...? */     case 5:         the_new_feature();         break; } 

It's common-mistakes in C/C++. If you add new feature after 5 years and you completely forget it, it'll become a very buggy bug. Some modern language (e.g. C#, ...) even forbid switch-case without break or return.

Conclusion: There's no problem in syntax, but it's very bad practice and using break is highly recommended.