c++ - C++ 后缀计算器不能连续使用 2 个操作数

Question

您好我已经使用向量（这是必需的）编写了一个后缀计算器并且遇到了麻烦。当我连续输入两个操作数时，它不会给出正确的答案。例如，“5 4 + 3 10 * +”在应该给出 39 时给出了答案“36”。我明白为什么它不起作用我只是想不出一种方法来处理这种情况。有人可以帮帮我吗？代码：

#include <iostream>
#include <sstream>
#include <string>
#include <vector>
using namespace std;

//splits the string into different parts seperated by space and stores in tokens
void SplitString(string s, char delim, vector<string> &tokens)
{
stringstream ss;
ss.str(s);
string item;
while(getline(ss, item, delim))
{
    tokens.push_back(item);
}
}

//preforms the operation denoted by the operand 
void operation(string operand, vector<int> &eqVec)
{
int temp1 = eqVec.at(0);
int temp2 = eqVec.at(1);

if(operand == "+")
{
    eqVec.push_back(temp1 + temp2);

}
else if(operand == "-")
{
    eqVec.push_back(temp1 - temp2);
}
else if(operand == "*")
{
    eqVec.push_back(temp1 * temp2);
}
else if(operand == "/")
{
    eqVec.push_back(temp1 / temp2);
}

}

int main()
{
const char DELIM = ' ';
int total;
string eq;

vector <int> eqVec;
vector<string> tokens;


cout<<"Welcome to the postfix calculator! " << endl;
cout<<"Please enter your equation: ";

//gets the input and splits into tokens
getline(cin, eq);
SplitString(eq, DELIM, tokens); 

//cycles through tokens
for(int i = 0; i < tokens.size(); ++i)
{

    //calls operation when an operand is encountered
    if(tokens.at(i) == "+" || tokens.at(i) == "-" || tokens.at(i) == "*" || tokens.at(i) == "/")
    {
        operation(tokens.at(i), eqVec);
    }

    //otherwise, stores the number into next slot of eqVec
    else
    {

        //turns tokens into an int to put in eqVec
        int temp = stoi(tokens.at(i));
        eqVec.push_back(temp);
    }

}

//prints out only remaining variable in eqVec, the total
cout<<"The answer is: " << eqVec.at(0) << endl;






return 0;
}

Answer 1

在休息并回来后，我找到了解决它的方法。在这里发帖以防将来有人遇到同样的问题。下面的代码块现在位于一系列 if 语句之前的操作函数的开头。

int temp1, temp2;
int size = eqVec.size();
if(size > 2)
{
    temp1 = eqVec.at(size - 1);
    temp2 = eqVec.at(size - 2);
    eqVec.pop_back();
    eqVec.pop_back();
}
else
{
    temp1 = eqVec.at(0);
    temp2 = eqVec.at(1);
    eqVec.pop_back();
    eqVec.pop_back();
}