C中的有限状态机

时间:2018-05-04 00:33:57

标签: c enums fsm

我试图在C中创建一个遵循此图的FSM: FSM

并且输出看起来像这样: enter image description here

我首先定义了一个包含所有可能状态的enum

typedef enum {
Start = 0,
Build_Id = 1,
Identifier = 2,
Build_Num = 3,
Number = 4,
Error = 5
} State_type;

这是我目前的代码:

State_type analyzeData(State_type* currState, char c);

int main(int argc, char** argv) {
  State_type currState = 0;
  for (int i = 1; i < strlen(*argv); ++i) {
    analyzeData(&currState, *argv[i]);
  }
}



State_type analyzeData(State_type* currState, char c) {
  if (currState == 0) {
    if (isblank(c)) {
      *currState = (State_type) 0;
      return *currState;
    }
    else if (isdigit(c)) {
      *currState = (State_type) 3;
      return *currState;
    }
    else if (isalpha(c)) {
      *currState = (State_type) 1;
      return *currState;
    }
  }
}

我的计划是基本上为所有其他可能状态使用一系列if-else语句。我想我是否正确地接近这一点有点困惑。我一直试图阅读其他FSM问题的答案,但没有任何意义。有人能指出我正确的方向吗?

2 个答案:

答案 0 :(得分:3)

你定义一个列出你的状态的枚举 - 好!

typedef enum {
    Start_state = 0,
    Build_Id_state = 1,
    Identifier_state = 2,
    Build_Num_state = 3,
    Number_state = 4,
    Error_state = 5
} State_type;

稍微更改您的州过渡代码

int
main(int argc, char** argv) {
  State_type currState = 0;
  Action_t action;
  char* p = *argv; char symbol;
  int len = strlen(p);
  //C-strings are zero-indexed
  for (int i=0; i < len; ++i) {
    action = analyzeData(&currState, classify(symbol=*p++));
    switch(action) {
        case None_act: break;
        case Gather_act: //appropriate symbol gathering
        case Emit_act: //handle ident/number print/save
        case Stop_act: //appropriate behavior, e.g. i=len
        ...
    }
  }
}

构建一个包含这些条目的状态转换表:

typedef struct state_table_entry_s {
    State_type state;
    Transition_t trans; //could define as bit-field
    State_type nextstate;
    Action_t action; //semantic action
} state_table_entry_t;

定义状态转换表,清楚表明您尚未定义某些转换的行为。 (使表格成为二维的,您可以更快地处理状态和转换)

state_table_entry_t states[] = {
    {Start_state, Letter_class, None_act, Build_Id}
   ,{Start_state, Digit_class, None_act, Build_Num}
   ,{Start_state, Blank_class, None_act, Start_state}
   ,{Start_state, Semicolon_class, Stop_act, Start_state}
   ,{Build_Id_state, Letter_class, Gather_act, Build_Id_state}
   ,{Build_Id_state, Digit_class, Gather_act, Build_Id_state}
   ,{Build_Id_state, Underscore_class, Gather_act, Build_Id_state}
   ,{Build_Id_state, Blank_class, None_act, Identifier_state}
   ,{Identifier_state, Blank_class, Emit_act, Start_state}
   ,{Build_Num_state, Digit_class, Gather_act, Build_Num_state}
   ,{Build_Num_state, Blank_class, None_act, Number_state}
   ,{Number_state, Blank_class, Emit_act, Start_state}
   ,{Stop_state, <any>, Error_act, Stop_state}
   ,{Error_state, <any>, None_act, Stop_state}
};

注意上面的“状态转换表”如何清楚地记录您的状态机?您可以(轻松地)从配置文件中加载此表吗?

停止。您是否为每个(状态X转换)对定义了(适当的)操作?

//States:
Start_state
Build_Id_state
Identifier_state
Build_Num_state
Number_state
Error_state

//Transitions:
Letter_class
Digit_class
Underscore_class
Blank_class
Semicolon_class
Other_class

对于上述内容,您需要定义状态转换类:

typedef enum {
    Letter_class
   ,Digit_class
   ,Underscore_class
   ,Blank_class
   ,Semicolon_class
   ,Other_class
} Transition_t;

你需要定义你的行动:

typedef enum {
    None_act
   ,Gather_act
   ,Emit_act
   ,Stop_act
   ,Error_act
} Action_t;

将您的字符/符号转换为其转换类(您可以使用ctype.h和isalpha(),isdigit()宏),

Transition_t classify(char symbol) {
    Transition_t class = Other_class;
    if (isblank(c)) {
      return(class = Blank_class); break;
    }
    else if(isdigit(symbol)) {
      return(class = Digit_class);
    }
    else if (isalpha(symbol)) {
      return(class = Letter_class); break;
    }
    else {
      switch(tolower(symbol)) {
        case ' ':
            return(class = Blank_class); break;
        case '_':
            return(class = Underscore_class); break;
        case ';':
            return(class = Semicolon_class); break;
        default :
            return(class = Other_class); break;
      }
    }
    return(class = Other_class); break;
}

在状态表中找到匹配状态(可以使效率更高),并在转换表中进行匹配转换,然后采取语义操作,

Action_t
analyzeData(State_type& currState, Transition_t class) {
  for( int ndx=0; ndx<sizeof(states)/sizeof(states[0]); ++ndx ) {
    if( (states[ndx].state == currState)
    &&. (states[ndx].trans == class) ) { //state match
        semantic_action(states[ndx].action);
        currState = states[ndx].nextState;
        return(states[ndx].action);
    }
  }
}

您需要定义'semantic_action'功能,当然您需要“收集”您的输入,以便您可以在适当的操作时间执行输出。你的'emit_act'需要清理。

答案 1 :(得分:1)

使用switch语句(甚至转换表)可能会更好,但基本结构是相同的。

如果您使用枚举,请使用它。不要使用魔术数字。定义枚举的关键是能够使用有意义的名称而不是数字。

如果要返回新状态,使用输入输出参数绝对没有意义。使用原型

State_type analyzeData(State_type currState, char c);
/* Better would be int c. See below. */

然后典型的州案例可能是:

case Start:
  if (isblank(c)) return Start;
  else (isdigit(c)) return Build_Num;
  else (isalpha(c)) return Build_Id;
  else return Error;

另请注意,isalpha和朋友可以使用int,而不是char。如果char已签名(这是常见的)且值恰好为负数,则会导致未定义的行为。