c - switch 语句中的独立中断

Question

我偶然发现了这段代码，它按预期工作:

switch (ev->deviceType) {

    break;

    case DEVICE_TS1E0:
        //some code
        break;

    case DEVICE_TS1E3:
       //some code
       break;

    default:
        //some logging
        break;
}

现在，在开关的开头有一个孤独的 break;，这似乎没有任何效果。

是否有任何情况下 break; 会产生效果？

Answer 1

TL;DR 该 break 语句无效并且是死代码。控制永远不会到达那里。

C11 标准有一个类似案例的很好的例子，让我直接引用它。

来自第 §6.8.4.2/7 章，(强调我的)

EXAMPLE In the artificial program fragment

switch (expr)
{
    int i = 4;
    f(i);
case 0:
    i = 17;
    /* falls through into default code */
default:
    printf("%d\n", i);
}

the object whose identifier is i exists with automatic storage duration (within the block) but is never initialized, and thus if the controlling expression has a nonzero value, the call to the printf function will access an indeterminate value. Similarly, the call to the function f cannot be reached.