Question

以下是一个实现DFA（确定性有限自动机）的简单程序。但是，我的问题不涉及DFA。

#include<iostream>
using namespace std;

int main()
{
    char c;
    int state=1;
    while((c=cin.get())!='4')
    {
        switch(state)
        {
            case 1:
            if(c=='0')
            state=2;
            if(c=='1')
            state=1;
            if(c=='2')
            state=2;

            break;

            case 2:
            if(c=='0')
            state=5;
            if(c=='1')
            state=1;
            if(c=='2')
            state=3;

            break;

            case 3:
            if(c=='0')
            state=1;
            if(c=='1')
            state=5;
            if(c=='2')
            state=4;

            break;

            case 4:
            if(c=='0')
            state=3;
            if(c=='1')
            state=4;
            if(c=='2')
            state=5;

            break;

            case 5:
            if(c=='0')
            state=5;
            if(c=='1')
            state=4;
            if(c=='2')
            state=1;

            break;

            default:
            cout<<"Input will not be accepted"<<endl;

        } //switch
        cout<<"Current state is "<<state<<endl; 
    } //while


    return 0;
}

当我运行代码时，我发现每行输出两次。例如，当我输入0 1 0 0 4时，DFA从状态1-> 2-> 1＆gt; 2-> 5开始，因此输出应为：

Current state is 2
Current state is 1
Current state is 2
Current state is 5

但输出是：

Current state is 2
Current state is 2
Current state is 1
Current state is 1
Current state is 2
Current state is 2
Current state is 5
Current state is 5

有谁可以指出原因？

Answer 1

cin.get()读取一个字符，因此您也在阅读空格。然后在每个空格之后，程序只输出先前的状态，因为空间不匹配任何东西。您想要使用cin >> c。

Answer 2

由于您显然不会将您的代码发布到CodeReview，所以让我发一个我认为我会编写代码的方法：

#include<iostream>
using namespace std;

static const int states[][3] = {
    { 2, 1, 2 },
    { 5, 1, 3 },
    { 1, 5, 4 },
    { 3, 4, 5 },
    { 5, 4, 1 }
};

int main()
{
    char c;
    int state=1;
    while(cin >> c && c != '4') {
        if (c >= '0' && c <= '2')
            state = states[state-1][c-'0'];
        else
            std::cout << "Input will not be accepted.\n";
        cout<<"Current state is "<<state<<"\n"; 
    }
    return 0;
}

至少在我看来，这似乎更容易阅读和理解，更不用说将状态机本身与读取字符的代码分开，并根据输入遍历状态。

输出中的冗余

2 个答案: