Postfix转换括号的中缀错误

Question

我已经编写了将中缀表达式转换为后缀表达式的代码。 有两个用于检查表达式的函数，一个用于检查不带括号的表达式，另一个用于检查和转换带括号的表达式。

转换非括号表达式的函数可以正常工作，但是使用括号转换的函数却出错。

当我输入和（a + b）*（cd）之类的表达式时，它将正确转换a + b，然后将*推入堆栈。 但是不知何故，程序无法将（cd）的起始括号识别为括号，并像普通情况一样继续转换（对于不带括号的情况）。 我不知道为什么它不承认它是一个括号，这个小问题毁了整个输出。

谁能告诉我为什么它不能识别（cd中的）作为开括号吗？这是我的代码：

class Stack
  {
    .
    .
    .

   void push(char value) //push value onto top of stack
   {
      if(isFull()==true) //only possible when stack is not full
        cout<<endl<<"Sorry! Stack is already full, no more entries allowed!"<<endl;
      else //add value to top of stack is stack is not empty
      {
        top++;
        array[top]=value;
        cout<<endl<<value<<" pushed onto the stack!"<<endl;
      }
   }

   char pop() //remove an element from the top
   {
       if(isEmpty()==true) //if stack empty, nothing can be popped
        cout<<endl<<"Sorry stack empty! Nothing to be popped!"<<endl;
       else // otherwise return element at located at the top
         {
           return array[top--];
           cout<<endl<<array[top]<<" popped"<<endl;
         }
   }

   bool checkPrecedence(char a, char b) // function to check the precedence of operators
   {
       int precA,precB;

       switch(a) //assigning value for precedence to operator A
       {
       case '+':
        precA=1;
        break;
       case '-':
        precA=1;
        break;
       case '*':
        precA=2;
        break;
       case '/':
        precA=2;
        break;
       case '^':
        precA=3;
        break;
       case '(':
        precA=0;
        break;
       }

       switch(b)  //assigning value for precedence to operator A
       {
       case '+':
        precB=1;
        break;
       case '-':
        precB=1;
        break;
       case '*':
        precB=2;
        break;
       case '/':
        precB=2;
        break;
       case '^':
        precB=3;
        break;
       case '(':
        precB=0;
        break;
       }

       if (precA<precB) //comparing precedence
        return false;
       else if (precA>precB)
        return true;
       else if (precA==precB)
        return true;
   }

   void checkExpression(char expr[]) //function to convert expression without brackets
   {
       int i=0; //control character number of input
       int j=0;  //control character number of output
       int iter=0;  //keep track of iterations
       int noStack=0; //to keep track of size of stack at a particular time
       char ch=expr[i]; //taking first character from input
       char c;

       noStack=top+1;


       cout<<endl<<strlen(expr)<<endl; //displaying length of expression

       while(i<strlen(expr)) //keep checking until the whole expression is not checked
       {
           iter++;
           if((ch=='+' || ch=='-' || ch=='*' || ch=='/' || ch=='^') && (isEmpty()==true)) // if we encounter an operator and stack is empty
            {
                 push(ch);  //push operator onto stack
                 noStack++;
                 cout<<endl<<retrieve_top();
            }

           if((noStack==1) && (iter==2)) //so that single operator in stack is not compared with itself and added to output
           {

           }

           else if(ch=='+' || ch=='-' || ch=='*' || ch=='/' || ch=='^') //if we encounter an operator when stack already contains an operator
           {
             cout<<endl<<retrieve_top();
             c=pop(); //pop the element on top
             noStack--;

             if( checkPrecedence(c,ch)== true) // and then compare the precedence of the 2
             {
               outputString[j]=c; //if top of stack has greater precedence, add it to output
               j++;
               cout<<endl<<c<<" added to output";
               push(ch); //and push the other operator onto stack
               noStack++;
               if(noStack==2)
               {
                   c=pop();
                   noStack--;
                   outputString[j]=c;
                   j++;
               }
               cout<<endl<<retrieve_top()<<endl;
             }

             else //otherwise first push the operator on stack which was previously on top and then push the other operator on top of it
             {
                 push(c);
                 noStack++;
                 push(ch);
                 noStack++;

                 cout<<endl<<retrieve_top()<<endl;
             }

           }

           else // otherwise, if operand encountered, add it to output
            {
                cout<<endl<<ch<<" added to output";
                outputString[j]=ch;
                j++;
            }
          i++;
          ch=expr[i]; //check next character from expression
       }

       if( (isEmpty()==false) && (i==strlen(expr)) ) //if input has ended and stack still contains operators
       {
           while(isEmpty()!=true) //keep popping and adding them to output until stack becomes empty
           {
               cout<<endl<<retrieve_top()<<endl;
               c=pop();
               noStack--;
               cout<<endl<<retrieve_top();
               cout<<endl<<c<<" added to output";
               outputString[j]=c;
               j++;
           }
       }
   }

   void checkBracedExpression(char expr[]) //function to convert expression containing brackets
   {

       int i=0;   //control character number of input
       int j=0;   //control character number of output
       int iter=0;  //keep track of iterations
       int noStack=0; //to keep track of size of stack at a particular time
       char ch=expr[i];  //taking first character from input
       char c;

       noStack=top+1;


       cout<<endl<<strlen(expr)<<endl;

       while(i<strlen(expr))
       {
           iter++;

           if (ch=='(') //if we encounter a bracket
           {
               do
               {

                   if (ch=='+' || ch=='-' || ch=='*' || ch=='/' || ch=='^'||ch=='(' ) //and character input is an operator, push it on the stack
                    {

                      push(ch);
                      noStack++;
                    }
                   else  //otherwise, add it to output
                   {
                       outputString[j]=ch;
                        j++;
                        cout<<endl<<ch<<" added to output";
                   }
                 i++;
                 ch=expr[i]; //check next character
               }
               while (ch!=')'); //keep checking until we reach )

               if(ch==')')  //when we reach )
               {
                   do //keep popping elements until we encounter ( in the stack
                   {
                       c=pop();
                       noStack--;
                       if (c=='(') //if ( is encountered, ignore it
                        {}
                       else
                       {
                           outputString[j]=c; //otherwise, add it to output
                           j++;
                           cout<<endl<<c<<" added to output";
                       }

                   }
                   while(c!='(');
               }

           }

           else // otherwise, continue normal expression checking
           {
             if((ch=='+' || ch=='-' || ch=='*' || ch=='/' || ch=='^') && (isEmpty()==true))  // if we encounter an operator and stack is empty
            {
                 push(ch);
                 noStack++;
            }

           if((noStack==1) && (iter==2)) //so that single operator in stack is not compared with itself and added to output
           {

           }

           else if(ch=='+' || ch=='-' || ch=='*' || ch=='/' || ch=='^')  //if we encounter an operator when stack already contains an operator
           {
             //cout<<endl<<retrieve_top();
             c=pop();
             noStack--;

             if( checkPrecedence(c,ch)== true)
              {
                 outputString[j]=c;
                 j++;
                 cout<<endl<<c<<" added to output";
                 push(ch);
                 noStack++;
                if(noStack==2)
                 {
                   c=pop();
                   noStack--;
                   outputString[j]=c;
                   j++;
                 }

              }

                else
                {
                  push(c);
                  noStack++;
                  push(ch);
                  noStack++;
                }

           }

           else
           {
            cout<<endl<<ch<<" added to output";
            outputString[j]=ch;
            j++;
           }
          i++;
          ch=expr[i];
       }


         i++;
         ch=expr[i];

       if( (isEmpty()==false) && (i==strlen(expr)) )
       {
           while(isEmpty()!=true)
           {
               c=pop();
               noStack--;
               cout<<endl<<c<<" added to output";
               outputString[j]=c;
               j++;
           }
       }


    }
}

   void displayOutputString()
   {

       cout<<endl<<"Output String: "<<outputString<<endl;
   }

  .
  .
  .


};

int main()
{
    char expression[50];
    Stack object;
    int option;

    object.createStack();

       do
    {
        cout<<endl<<"1.Convert simple expression."<<endl<<"2.Convert bracket expression."<<endl<<"3.Exit"<<endl<<"Your option:  ";
        cin>>option;
        cout<<endl;
        switch(option)
        {
        case 1:
          cout<<endl<<"Enter the expression (in infix): ";
          cin.ignore();
          cin>>expression;
          object.checkExpression(expression);
          object.displayOutputString();

          break;
        case 2:
          cout<<endl<<"Enter the expression (in infix): ";
          cin.ignore();
          cin>>expression;
          object.checkBracedExpression(expression);
          object.displayOutputString();

          break;
        }
    }
    while (option!=3);

     return 0;

}

Answer 1

在checkBracedExpression（）中

while(c!='(');

您可以尝试以下代码吗？

i++;
ch=expr[i]; //check next character
continue;

看看是否可行？