在案例中更改开关变量 [英] Changing switch variable inside a case

问题描述

在以下代码中:

int i = 0;

switch(i)
{
    case 0:
        cout << "In 0" << endl;
        i = 1;
        break;
    case 1:
        cout << "In 1" << endl;
        break;
}

会发生什么?它会调用未定义的行为吗?

What will happen? Will it invoke undefined behavior?

推荐答案

没有未定义的行为.但是，仅当代码到达 switch(i)时才测试 i 的值.因此，情况1:将被跳过(通过 break; 语句).

No undefined behavior. But the value of i is only tested when the code reaches switch (i). So case 1: will be skipped (by the break; statement).

switch 关键字并不意味着只要 i 的值为0/1，就运行代码".这意味着，请检查 i 现在是什么，然后基于该代码运行代码.不管将来 i 会发生什么.

The switch keyword does not mean "run code whenever the value of i is 0 / 1". It means, check what i is RIGHT NOW and run code based on that. It doesn't care what happens to i in the future.

实际上，有时这样做很有用:

In fact, it's sometimes useful to do:

for( step = 0; !cancelled; ++step ) {
    switch (step)
    {
        case 0:
            //start processing;
            break;

        case 1:
            // more processing;
            break;

        case 19:
            // all done
            return;
    }
}

在构建有限状态机时，在 case 块内更改控制变量非常普遍(尽管不是必需的，因为您可以在内设置 next_state 情况，然后再分配 state = next_state ).

And changing the control variable inside a case block is extremely common when building a finite state machine (although not required, because you could set next_state inside the case, and do the assignment state = next_state afterward).

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