c++ - 根据一些规则随机选择枚举值

Question

请看下面的代码

#include <iostream>
#include <stdlib.h>

using namespace std;

int main()
{
    enum Movement{STAND,WALK,RUN,CRAWL};

    Movement state = (Movement)(1+rand()%4);

    for(int i=0;i<100;i++)
    {

    cout << state << endl;

    switch(state)
    {
        case STAND: 
            cout << "You can walk or crawl" << endl;        
            while(state==WALK || state==CRAWL){
            state = (Movement)(1+rand()%4);}
            break;


        case WALK: 
            cout << "You can stand or run" << endl;
            while(state==STAND || state==RUN){
            state = (Movement)(1+rand()%4);}
            break;


        case RUN: 
            cout << "You can walk" << endl;
            while(state==WALK){
            state = (Movement)(1+rand()%4);}
            break;

        default: 
            cout << "You can stand" << endl;
            while(state==STAND){
            state = (Movement)(1+rand()%4);}

    }

    }
}

在这里，几乎没有规则。让我们称它们为合法过渡

1. 站立，他可以走路或爬行
2. 从步行开始，他可以站立或运行
3. 从运行开始，他可以走路
4. 从爬行开始，他可以站立

所以，在这里，如果合法的转换应该是随机选择的。但是，但是，它应该根据上面提供的规则。例如，如果选择了“STAND”，那么接下来要选择的应该是“WALK”或“CRAWL”。但是正如你在这里看到的，所有的结果

2
You can walk

这是为什么呢？请帮忙!

更新:

下面的代码是do..while循环，如回复中所建议的

#include <iostream>
#include <stdlib.h>

using namespace std;

int main()
{
    enum Movement{STAND,WALK,RUN,CRAWL};

    Movement state = (Movement)(1+rand()%4);

    for(int i=0;i<10;i++)
    {

    cout << state;

    if(state==STAND)
    {
        cout << " STAND is selected" << endl;

    }
    else if(state==WALK)
    {
        cout << " WALK is selected" << endl;

    }
    else if(state==RUN)
    {
        cout << " RUN is selected" << endl;
    }
    else if(state==CRAWL)
    {
        cout << " CRAWL is selected" << endl;
    }

    switch(state)
    {
        case STAND: 
            //cout << "You can walk or crawl" << endl;        
            do{state = (Movement)(rand()%4);}
            while(state==WALK || state==CRAWL);

            break;


        case WALK: 
            //cout << "You can stand or run" << endl;
            do{state = (Movement)(rand()%4);}
            while(state==STAND || state==RUN);
            break;


        case RUN: 
            //cout << "You can walk" << endl;
            do{state = (Movement)(rand()%4);}
            while(state==WALK);
            break;

        default: 
            //cout << "You can stand" << endl;
            do{state = (Movement)(rand()%4);}
            while(state==STAND);

    }

    }
}

还是一样。现在我得到了不同的答案，但不是正确的答案!!

Answer 1

你的逻辑顺序错误；您的 while 循环将全部被跳过，因为 state 仍然是您进入特定 case 时的状态。您可以使用 do-while 重写它们以确保它们至少运行一次。