c - 我们应该打破 switch 语句中的 default case 吗？

Question

假设此示例代码 ( 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;
}

我们应该在 last default 情况下使用 break; 吗？如果我删除它，我会看到程序的相同行为。但是，我看到其他示例也在 default 情况下使用了 break;。

为什么？有什么原因吗？

Answer 1

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

来自 C 编程语言 - 第二版 (K&R 2):

第3.4章开关

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.