Shunting Yard与c ++伪代码中数学表达式的评价 [英] Shunting Yard and Evaluation of mathematical expression in c++ pseudo code

问题描述

I am trying to transform a infix mathematical expression into a postfix expression using the shunting yard algortihm but can only get so far by writing the pseudo code because I know exactly what to do through logical thinking but can't write the code that will help me do it. If you can have a look at the code I wrote and help me in writing the code, I will really appreciate it.

Same with evaluating/calculating that postFix that was transformed. I also wrote the pseudo code for that function with the hope that someone will be able to help me.

This is my pseodo code:

Shunting Yard:

string shuntingYard(string expression)
{

    string postFix = "";    //An initially empty array
    int ops[];              //Array ops
    int i = -1;             //This will be an index into the array ops

    while(/*tokens must still be read*/){

        //Read next token from the expression

        if(/*if the token is a constant or a variable*/)
        {
            //Append the token to the back of the postFix string
        }
        else if(/*the token is a function*/)
        {
            if(i == -1)
            {
                ++i;
                ops[i] = token;
            }
        }
        else if(/*the token ia a operator*/)
        {
            if(i == -1)
            {
                ++i;
                ops[i] = token;
            }
            else
            {
                while(/*i > -1 && ops[i] is an operator && token is left-associative && its precedence <= to that of the operator at ops[i] or the token's precedence is < that of the operator at ops[i]*/)
                {
                    //append ops[i] to the back of the postFix string;
                    --i;
                }
                if(i == -1)
                {
                    ++i;
                }
                ops[i] = token;     //Place the operator at index i in ops
            }
        }
        else if(/*token is a left-parenthesis*/)
        {
            if(i == -1)
            {
                ++i;
            }
            //place the token at ops[i];
        }
        else    //The token must be a right parenthesis
        {
            while(/*ops[i] is NOT a left parenthesis*/)
            {
                //append ops[i] to the back of the postFix string
                --i;
            }
            //ops[i] at this point should be a left parenthesis
            --i;
            if(/*ops[i] is a function, append it to the postFix string*/)
            {
                --i;
            }
        }
    }
    while(i > -1)
    {
        //append ops[i] to the back of the postFix string;
        --i;
    }
    //return the postFix string;
}

这是对后缀表达式的评估/计算：

and this is the evaluation/calculation of the postfix expression:

int evaluate(string expr)
{

    int operands[];
    int i = -1;

    while(/*expr still has tokens to read*/)
    {
        /*token = next token from expr;*/
        if(/*If token is an integer or a variable*/){
            operands[++i] = token;
        }
        else
        {
            if(/*Token is a binary operator*/)
            {
                //Retrieve the elements from the operands at indices i and i-1;

                //Apply the operator to these operands;

                //Place the answer at operands[i-1];

                --i;
            }
            else    //The token is a unary operator
            {
                //Retrieve the element at index i from operands;

                //Apply the unary operator to this element;

                //Place the answer at operands[i];
            }
        }
    }
    //The answer for the expression will be at operands[0];
}

非常感谢你！



：）

Thank you very very much!

:)

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