使用堆栈将中缀表达式转换为后缀（C）

Question

我的讲师给了我一个任务，让我创建一个程序，使用堆栈将表达式和中缀转换为后缀。我制作了堆栈类和一些函数来读取中缀表达式。

但是这个函数，叫做transtToPostfix（char*const inFix，char*const postFix），它负责使用堆栈将数组inFix中的inFix表达式转换为数组postFix中的postfix表达式，并没有做它应该做的事情...你们能帮帮我告诉我哪里做错了吗？

下面是从中缀转换为后缀的函数的代码，convertToPostfix（char*const-inFix，char*const-postFix）是我需要帮助修复的代码：

 void ArithmeticExpression::inputAndConvertToPostfix()
    {
       char inputChar; //declaring inputChar
       int i = 0; //inizalize i to 0

       cout << "Enter the Arithmetic Expression(No Spaces): ";

       while( ( inputChar = static_cast<char>( cin.get() ) ) != '\n' )
       {
          if (i >= MAXSIZE) break; //exits program if i is greater than or equal to 100

          if(isdigit(inputChar) || isOperator(inputChar))
          {
             inFix[i] = inputChar; //copies each char to inFix array
             cout << inFix[i] << endl;
          }
          else
             cout << "You entered an invalid Arithmetic Expression\n\n" ;

          }

          // increment i;
          i++;
          convertToPostfix(inFix, postFix);


       }




    bool ArithmeticExpression::isOperator(char currentChar)
    {

        if(currentChar == '+')
            return true;
        else if(currentChar == '-')
            return true;
        else if(currentChar == '*')
            return true;
        else if(currentChar == '/')
            return true;
        else if(currentChar == '^')
            return true;
        else if(currentChar == '%')
            return true;
        else
            return false;
    }

    bool ArithmeticExpression::precedence(char operator1, char operator2)
    {
        if ( operator1 == '^' )
           return true;
        else if ( operator2 == '^' )
           return false;
        else if ( operator1 == '*' || operator1 == '/' )
           return true;
        else if ( operator1 == '+' || operator1 == '-' )
           if ( operator2 == '*' || operator2 == '/' )
              return false;
           else
              return true;

        return false;
    }

   void ArithmeticExpression::convertToPostfix(char * const inFix, char * const postFix)
        {
           Stack2<char> stack;

           const char lp = '(';

           stack.push(lp); //Push a left parenthesis ‘(‘ onto the stack.

           strcat(inFix,")");//Appends a right parenthesis ‘)’ to the end of infix.

          // int i = 0;
           int j = 0;

           if(!stack.isEmpty())
           {

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

                   if(isdigit(inFix[i]))
                   {
                        postFix[j] = inFix[i];
                        cout << "This is Post Fix for the first If: " << postFix[j] << endl;
                        i++;
                        j++;
                   }

                    if(inFix[i] == '(')
                   {
                       stack.push(inFix[i]);
                       cout << "The InFix was a (" << endl;
                       i++;
                       //j++;
                   }

                    if(isOperator(inFix[i]))
                               {
                            char operator1 = inFix[i];

                            cout << "CUrrent inFix is a operator" << endl;
                                   if(isOperator(stack.getTopPtr()->getData()))
                                       {
                                       cout << "The stack top ptr is a operator1" << endl;
                                       char operator2 = stack.getTopPtr()->getData();
                                           if(precedence(operator1,operator2))
                                           {
                                               //if(isOperator(stack.getTopPtr()->getData())){
                                                   cout << "The stack top ptr is a operato2" << endl;
                                                   postFix[j] = stack.pop();
                                                   cout << "this is post fix " << postFix[j] << endl;
                                                   i++;
                                                   j++;
                                              // }

                                           }

                                       }
                                   else

                                       stack.push(inFix[i]);
                                   // cout << "Top Ptr is a: "<< stack.getTopPtr()->getData() << endl;



                               }

                    for(int r = 0;r != '\0';r++)
                        cout << postFix[r] << " ";

                        if(inFix[i] == ')')
                       {
                           while(stack.stackTop()!= '(')
                         {
                               postFix[j] = stack.pop();
                               i++;
                               j++;
                                }
                           stack.pop();

                            }
                       }
           }

                   }

注意：convertToPostfix函数是使用以下算法生成的：

>

>

如果内缀中的当前字符是数字，则将其复制到后缀的下一个元素。

如果中缀的当前字符是左括号，将其推到堆栈上。

如果内缀中的当前字符是运算符，

• （如果在堆栈顶部有Pop运算符，则它们的优先级等于或高于Pop运算符的优先级）

• 从堆栈顶部弹出运算符并将其插入后缀，直到左括号位于堆栈顶部
• 从堆栈中弹出（并放弃）左括号

Answer 1

这是我用C和多个数字的评估。

#include <stdio.h>
#include <math.h>
#define MAX 50
void push(char[],char);
void in_push(double[], double);
int pop();
int prec(char);
double eval(char[],int,double[]);
int top = 0;
void main() {
double eval_stack[MAX];
int op_count=0;
char stack[MAX], exps[MAX], symbols[MAX];
int i=0,j=0,len,check;
while((symbols[i]=getchar())!='\n') {
if(symbols[i]!=' ' || symbols[i]!='\t') {
if(symbols[i]=='+' || symbols[i]=='-' || symbols[i]=='/' || symbols[i]=='*' || symbols[i]=='^')
op_count++;
i++;
}
}
symbols[i]='#';
symbols[++i]='\0';
len = strlen(symbols);
stack[top] = '#';
for(i=0; i<=len; i++) {
if(symbols[i]>='a'  && symbols[i]<='z') {
exps[j]=symbols[i];
j++;
}
switch(symbols[i]) {
case '0': case '1': case '2': case '3': case '4': case '5': case '6': case '7': case '8': case '9':
//if(symbols[i]>='a'  && symbols[i]<='z') {
exps[j]=symbols[i];
j++;
break;
case '+': case '-': case '*': case '/': case '^':
exps[j++] = ' ';
while(prec(symbols[i]) <= prec(stack[top])) {

             exps[j] = stack[top];
             pop();
             //printf("\n\t\t%d\t\t%d\n", top,j);
                j++;


}
if(prec(symbols[i]) > prec(stack[top])) {
push(stack,symbols[i]);
}
break;
case '(':
push(stack,symbols[i]);
break;
case ')':
while(stack[top]!='(') {
      exps[j] = stack[top];
       pop();
      j++;

}
pop();
break;
case '#':
exps[j++] = ' ';
while(stack[top]!='#') {
      exps[j] = stack[top];
       pop();
       j++;

}
pop();
break;
}
}
exps[j]='\0';
printf("Postfix: %s", exps);
for(i=0; i<j; i++)
if(exps[i]=='a')
check = 1;
if(check!=1)
printf("\nSolution: %.1f", eval(exps,j,eval_stack));
}

double eval(char exps[],int exps_len,double eval_stack[]) {
    int i; int len=exps_len,temp;
    double in_temp[MAX],t;
    int count,power,j,in_count;
    count=power=j=t=in_count=0;
    double result;
for(i=0; i<len; i++) {
switch(exps[i]) {
case '0': case '1': case '2': case '3': case '4': case '5': case '6': case '7': case '8': case '9':
in_temp[i] = exps[i]-'0';
j=i+1;
while(exps[j]>='0' && exps[j]<='9') {
in_temp[j] = exps[j]-'0';
j++; // 2
}
count = i; // 3
while(in_temp[count]<='0' && in_temp[count]<='9') {
power = (j-count)-1;
t = t + in_temp[count]*(pow(10,power));
power--;
count++;
}
in_push(eval_stack,t);
i=j-1;
t=0;
break;
case '+':
temp = pop();
pop();
result = eval_stack[temp] + eval_stack[temp+1];
in_push(eval_stack,result);
break;
case '-':
temp = pop();
pop();
result = eval_stack[temp] - eval_stack[temp+1];
in_push(eval_stack,result);
break;
case '*':
temp = pop();
pop();
result = eval_stack[temp] * eval_stack[temp+1];
in_push(eval_stack,result);
break;
case '/':
temp = pop();
pop();
result = eval_stack[temp] / eval_stack[temp+1];
in_push(eval_stack,result);
break;
case '^':
temp = pop();
pop();
result = pow(eval_stack[temp],eval_stack[temp+1]);
in_push(eval_stack,result);
break;
}
}
return eval_stack[top];
}
int prec(char a) {
if(a=='^')
return 3;
else if(a=='*' || a=='/' || a=='%')
return 2;
else if(a=='+' || a=='-')
return 1;
else if(a=='(')
return 0;
else
return -1;
}

void push(char stack[], char ele) {
    if(top>=MAX) {
    printf("\nStack Overflow");
    exit(1);
    }
stack[++top] = ele;
}
void in_push(double stack[], double ele) {
    if(top>=MAX) {
    printf("\nStack Overflow");
    exit(1);
    }
stack[++top] = ele;
}
int pop() {
    if(top<0) {
    printf("\nStack Underflow");
    exit(1);
    }

top = top - 1;
return top;
}

Answer 2

因此，您的代码存在许多问题。我将发布（应该是）一个纠正的解决方案，其中有大量的评论来解释发生了什么以及你在哪里犯了错误。前面有几件事：

>

我正在使用标准库中的堆栈

它是一个自由函数，不是类的一部分，但应该很容易修改——这样测试对我来说很简单。

我不相信你的操作员优先级表是正确的，但是，我会让你仔细检查一下。

#include <stack>
#include <cctype>
#include <iostream>

std::string convertToPostfix(std::string& infix)
{
    std::string postfix; //Our return string
    std::stack<char> stack;
    stack.push('('); //Push a left parenthesis ‘(‘ onto the stack.
    infix.push_back(')');

    //We know we need to process every element in the string,
    //so let's do that instead of having to worry about
    //hardcoded numbers and i, j indecies
    for(std::size_t i = 0; i < infix.size(); ++i) {

        //If it's a digit, add it to the output
        //Also, if it's a space, add it to the output 
        //this makes it look a bit nicer
        if(isdigit(infix[i]) || isspace(infix[i])) {
            postfix.push_back(infix[i]);
        }

        //Already iterating over i, so 
        //don't need to worry about i++
        //Also, these options are all mutually exclusive,
        //so they should be else if instead of if.
        //(Mutually exclusive in that if something is a digit,
        //it can't be a parens or an operator or anything else).
        else if(infix[i] == '(') {
            stack.push(infix[i]);
        }

        //This is farily similar to your code, but cleaned up. 
        //With strings we can simply push_back instead of having
        //to worry about incrementing some counter.
        else if(isOperator(infix[i]))
        {
            char operator1 = infix[i];
            if(isOperator(stack.top())) {
                while(!stack.empty() && precedence(operator1,stack.top())) {
                    postfix.push_back(stack.top());
                    stack.pop();
                }
            }
            //This shouldn't be in an else - we always want to push the
            //operator onto the stack
            stack.push(operator1);
        }    

        //We've hit a right parens - Why you had a for loop
        //here originally I don't know
        else if(infix[i] == ')') {
            //While top of stack is not a right parens
            while(stack.top() != '(') {
            //Insert into postfix and pop the stack
                postfix.push_back(stack.top());
                stack.pop();
            }
            // Discard the left parens - you'd forgotten to do this
            stack.pop(); 
        }
    }

    //Remove any remaining operators from the stack
    while(!stack.empty()) {
        postfix.push_back(stack.top());
        stack.pop();
    }
}

Answer 3

这基本上是对Yuushi回答的评论。

• 外部while（！stack.empty（））循环错误。把它拿走。（保持循环体为ofc）。在函数末尾，检查堆栈是否为空，否则表达式有语法错误
• 正如Yuushi已经说过的，优先函数看起来是假的。首先，应该为参数指定更好的名称：一个是左边的运算符，另一个是右边的运算符。（现在称之为优先（rightOp，leftOp））。然后，您应该记录结果的含义-如果a rOp b lOp c==（a rOp b）lOp c（是的，运算符顺序与您所称的不匹配，例如-“”和“-”在两个顺序中都不相同，那么现在返回true）
• 如果您找到一个新运算符，则需要在堆栈上的旧运算符上循环，例如，在读取a-b*c后，您的输出是a-b-c，堆栈是[-*]。现在您阅读了一个，需要同时弹出两个操作符，结果是abc*-。即，输入a-b*cd应导致a-b*cd

更新：附加完整的解决方案（基于Yuushi的回答）：

bool isOperator(char currentChar)
{
    switch (currentChar) {
    case '+':
    case '-':
    case '*':
    case '/':
    case '^':
    case '%':
        return true;
    default:
        return false;
    }
}

// returns whether a `lOp` b `rOp` c == (a `lOp` b) `rOp` c
bool precedence(char leftOperator, char rightOperator)
{
    if ( leftOperator == '^' ) {
        return true;
    } else if ( rightOperator == '^' ) {
        return false;
    } else if ( leftOperator == '*' || leftOperator == '/' || leftOperator == '%' ) {
        return true;
    } else if ( rightOperator == '*' || rightOperator == '/' || rightOperator == '%' ) {
        return false;
    }

    return true;
}

#include <stdexcept>
#include <cctype>
#include <sstream>
#include <stack>
std::string convertToPostfix(const std::string& infix)
{
    std::stringstream postfix; // Our return string
    std::stack<char> stack;
    stack.push('('); // Push a left parenthesis ‘(‘ onto the stack.

    for(std::size_t i = 0, l = infix.size(); i < l; ++i) {
        const char current = infix[i];

        if (isspace(current)) {
            // ignore
        }
        // If it's a digit or '.' or a letter ("variables"), add it to the output
        else if(isalnum(current) || '.' == current) {
            postfix << current;
        }

        else if('(' == current) {
            stack.push(current);
        }

        else if(isOperator(current)) {
            char rightOperator = current;
            while(!stack.empty() && isOperator(stack.top()) && precedence(stack.top(), rightOperator)) {
                postfix << ' ' << stack.top();
                stack.pop();
            }
            postfix << ' ';
            stack.push(rightOperator);
        }

        // We've hit a right parens
        else if(')' == current) {
            // While top of stack is not a left parens
            while(!stack.empty() && '(' != stack.top()) {
                postfix << ' ' << stack.top();
                stack.pop();
            }
            if (stack.empty()) {
                throw std::runtime_error("missing left paren");
            }
            // Discard the left paren
            stack.pop();
            postfix << ' ';
        } else {
            throw std::runtime_error("invalid input character");
        }
    }


    // Started with a left paren, now close it:
    // While top of stack is not a left paren
    while(!stack.empty() && '(' != stack.top()) {
        postfix << ' ' << stack.top();
        stack.pop();
    }
    if (stack.empty()) {
        throw std::runtime_error("missing left paren");
    }
    // Discard the left paren
    stack.pop();

    // all open parens should be closed now -> empty stack
    if (!stack.empty()) {
        throw std::runtime_error("missing right paren");
    }

    return postfix.str();
}

#include <iostream>
#include <string>
int main()
{
    for (;;) {
        if (!std::cout.good()) break;
        std::cout << "Enter the Arithmetic Expression: ";
        std::string infix;
        std::getline(std::cin, infix);
        if (infix.empty()) break;

        std::cout << "Postfix: '" << convertToPostfix(infix) << "'\n";
    }

    return 0;
}