RPN评估C ++

Question

Hello）这是我从中缀表达式转换为后缀表达式的代码，但是我无法理解如何评估后缀表达式，我将非常感谢任何提示。我不是要求代码，尽管这会有所帮助。

#include <iostream> 
#include <stack> 
#include <string>

using namespace std; 
bool operation(char b) 
{
    return b=='+' || b=='-' || b=='*' || b=='/' ; 

} 

bool priority(char a, char b) 
{
    if(a=='(')
    {
        return true; 
    } 
    if(a=='+' || a=='-') 
    {
        return true; 

    } 
    if(b=='+' || b=='-') 
    {
        return false; 
    } 

    return true; 

} 
int main() 
{

    string a;
    string res; 
    stack<char> s; 
        cin>>a; 
    for(int i=0; i<a.size(); i++) 
    {
        if(a[i]!='(' && a[i]!=')' && operation(a[i])==false) 
        {

            res=res+a[i]; 
        } 
        if(a[i]=='(') 
        {
            s.push(a[i]) ; 
        } 
        if(a[i]==')') 
        {
            while(s.top()!='(') 
            {
                res+=s.top(); 
                s.pop();
            }
            s.pop(); 
        } 
        if(operation(a[i])==true) 
        {
            if(s.empty() || (s.empty()==false && priority(s.top(), a[i])) ) 
            {
                s.push(a[i]); 
            }
            else 
            {
                while(s.empty()==false && priority(s.top(),a[i])==false ) 
                { 
                    res+=s.top(); 
                    s.pop(); 
                }
                s.push(a[i]) ; 
            }

        } 

    } 
    while(s.empty()==false) 
    {
        res+=s.top(); 
        s.pop(); 
    } 
    cout<<res; 




    return 0; 
} 

P.S。我没有任何评论，但我认为代码本身是不言自明的）） P.P.S.先感谢您。

Answer 1

如果你形成用空格分隔的后缀表达式，下面将是对评估者进行编码的最简单方法之一，只是仅仅遵循评估的algorithm

这假设RPN类似5 1 2 + 4 * + 3 -（由空格分隔）

int evaluate_posfix ( const std::string& expression )
{

    int l,r,ans;
    std::stringstream postfix(expression);
    std::vector<int> temp;
    std::string s;
    while ( postfix >> s )
    {
        if( operation(s[0]) )
        {
            //Pull out top two elements
            r = temp.back();
            temp.pop_back();
            l = temp.back();
            temp.pop_back();
            // Perform the maths
            switch( s[0])
            {
                case '+': ans =  l + r ; break;
                case '-': ans =  l - r ; break;
                case '*': ans =  l * r ; break;
                case '/': ans =  l / r ; break; // check if r !=0
            }

            temp.push_back( ans ); // push the result of above operation
        }
        else
        {
            temp.push_back( std::stoi(s) );
        }
    }

    return temp[0] ; //last element is the answer
} 

Answer 2

尝试以下代码（在VS2019上测试）：

// CPP program to evaluate a given 
// expression where tokens are 
// separated by space. 
#include <string>
#include <stack>
#include <iostream>

// Function to find precedence of 
// operators. 
int precedence(char op) {
    if (op == '+' || op == '-')
        return 1;
    if (op == '*' || op == '/')
        return 2;
    return 0;
}

// Function to perform arithmetic operations. 
int applyOp(int a, int b, char op) {
    switch (op) {
    case '+': return a + b;
    case '-': return a - b;
    case '*': return a * b;
    case '/': return a / b;
    }
}

// Function that returns value of 
// expression after evaluation. 
int evaluate(std::string tokens) {
    int i;

    // stack to store integer values. 
    std::stack <int> values;

    // stack to store operators. 
    std::stack <char> ops;

    for (i = 0; i < tokens.length(); i++) {

        // Current token is a whitespace, 
        // skip it. 
        if (tokens[i] == ' ')
            continue;

        // Current token is an opening 
        // brace, push it to 'ops' 
        else if (tokens[i] == '(') {
            ops.push(tokens[i]);
        }

        // Current token is a number, push 
        // it to stack for numbers. 
        else if (isdigit(tokens[i])) {
            int val = 0;

            // There may be more than one 
            // digits in number. 
            while (i < tokens.length() &&
                isdigit(tokens[i]))
            {
                val = (val * 10) + (tokens[i] - '0');
                i++;
            }

            values.push(val);
        }

        // Closing brace encountered, solve 
        // entire brace. 
        else if (tokens[i] == ')')
        {
            while (!ops.empty() && ops.top() != '(')
            {
                int val2 = values.top();
                values.pop();

                int val1 = values.top();
                values.pop();

                char op = ops.top();
                ops.pop();

                values.push(applyOp(val1, val2, op));
            }

            // pop opening brace. 
            ops.pop();
        }

        // Current token is an operator. 
        else
        {
            // While top of 'ops' has same or greater 
            // precedence to current token, which 
            // is an operator. Apply operator on top 
            // of 'ops' to top two elements in values stack. 
            while (!ops.empty() && precedence(ops.top())
                >= precedence(tokens[i])) {
                int val2 = values.top();
                values.pop();

                int val1 = values.top();
                values.pop();

                char op = ops.top();
                ops.pop();

                values.push(applyOp(val1, val2, op));
            }

            // Push current token to 'ops'. 
            ops.push(tokens[i]);
        }
    }

    // Entire expression has been parsed at this 
    // point, apply remaining ops to remaining 
    // values. 
    while (!ops.empty()) {
        int val2 = values.top();
        values.pop();

        int val1 = values.top();
        values.pop();

        char op = ops.top();
        ops.pop();

        values.push(applyOp(val1, val2, op));
    }

    // Top of 'values' contains result, return it. 
    return values.top();
}

int main() {
    std::cout << evaluate("10 + 2 * 6") << "\n";
    std::cout << evaluate("100 * 2 + 12") << "\n";
    std::cout << evaluate("100 * ( 2 + 12 )") << "\n";
    std::cout << evaluate("100 * ( 2 + 12 ) / 14");
    return 0;
}

// This code is contributed by Nikhil jindal. 

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文章：Expression Evaluation