分流场算法C++方法之间未正确传递?

Question

我的任务是编写一个分流场算法，用于我的最终项目(计算器)。我已经以对我来说有意义的方式编写了程序，但是，在调用主算法函数 (toRPN) 时，我没有得到任何输出。我相信这是在解析和 toRPN 之间传递值的问题，因为我直接在 main 中测试了解析并且工作正常，但是当我尝试在 toRPN 函数中进行打印测试时，它什么也没打印。有人可以指出我正确的方向吗？

页眉：

#include <iostream>
#include <math.h>
#include <vector>
#include <stack>
#include <queue>
using namespace std;
#ifndef SHUNTING_YARD_ALGORITHM_SHUNTINGYARD_H
#define SHUNTING_YARD_ALGORITHM_SHUNTINGYARD_H

class ShuntingYard {
public:
stack <string> stack;
vector <string> tokens;
queue <string> outputList;
vector <char> operators;
vector <int> precedence;
vector <char> associativity;
ShuntingYard ();
bool hasOnlyDigits(const string s);
int getPrecedence(const string s);
int getAssociativity(const char c);
vector<string> parse(const string input) const;
string mainAlgorithm(const string);
};
#endif //SHUNTING_YARD_ALGORITHM_SHUNTINGYARD_H

.cpp：

#include "ShuntingYard.h"
#include <iostream>
#include <math.h>
#include <vector>
#include <stack>
#include <queue>
#include <sstream>
#include <numeric>
using namespace std;
stack <string> stack1;
queue <string> outputList;
vector <string> operators;
vector <int> precedence;
vector <char> associativity;
ShuntingYard::ShuntingYard () = default;
bool hasOnlyDigits(const string s){
return s.find_first_not_of( "0123456789" ) == string::npos;
}
int getPrecedence(const string s) {
for(int i = 0; i < operators.size(); i++) {
if (s ==  operators[i])
return precedence[i];
}
}
char getAssociativity(const string s) {
for(int i = 0; i < operators.size(); i++) {
if (s == operators[i])
return associativity[i];
}
}
vector<string> parse(const string input) {
// Parses the string by white space
istringstream ss(input);
vector <string> tokenVector;
// Fill vector with ss
for (string input; ss >> input;) {
tokenVector.push_back(input);
}
return tokenVector;
}

string toRPN(const string s) {
// Delimit string by white space and store in vector
vector <string> tokens = parse(s);
// Test print
for (int i = 0; i < tokens.size(); i ++)
cout << tokens[i];
//Change "rt" to "$" to be easily accessed
for (int i = 0; i < tokens.size(); i ++) {
if (tokens[i] == "rt")
tokens[i] = "$";
}
// Stores operators and their precedence/associativity to vectors using same index
operators.push_back("+"); precedence.push_back(2); associativity.push_back('L');
operators.push_back("-"); precedence.push_back(2); associativity.push_back('L');
operators.push_back("/"); precedence.push_back(3); associativity.push_back('L');
operators.push_back("*"); precedence.push_back(3); associativity.push_back('L');
operators.push_back("^"); precedence.push_back(4); associativity.push_back('R');
operators.push_back("$"); precedence.push_back(4); associativity.push_back('R');

// Shunting-Yard logic
while (tokens.size() != 0) {
for (int i = 0; i < tokens.size(); i++) {
if (hasOnlyDigits(tokens[i]))
outputList.push(tokens[i]);
if ( find(operators.begin(), operators.end(), tokens[i]) != operators.end()) {
while (getPrecedence(stack1.top()) > getPrecedence(tokens[i]) || (getPrecedence(stack1.top()) == getPrecedence(tokens[i]) &&
getAssociativity(tokens[i]) == 'L') && stack1.top() != "(") {
outputList.push(stack1.top());
stack1.pop();
stack1.push(tokens[i]);
}
}
if (tokens[i] == "(")
stack1.push(tokens[i]);
if (tokens[i] == ")")
while(!stack1.empty() && stack1.top() != "(") {
outputList.push(stack1.top());
stack1.pop();
}
stack1.pop();
}
if (tokens.size() == 0) {
while(!stack1.empty()) {
outputList.push(stack1.top());
stack1.pop();
}
}
}
// Replaces values with "$" back to "rt"
string str;
while (!outputList.empty()) {
if (outputList.front() == "$") {
str.insert(0,"rt");
outputList.pop();
}
else {
str.insert(0, (outputList.front()));
outputList.pop();
}
}
return str;
}
int main() {
string s1 = "3 + 4";
cout << toRPN(s1);
}

更新：

我已将问题缩小到以下 while 循环：

while (getPrecedence(stack1.top()) > getPrecedence(tokens[i]) || (getPrecedence(stack1.top()) == getPrecedence(tokens[i]) &&
getAssociativity(tokens[i]) == 'L') && stack1.top() != "(") {
outputList.push(stack1.top());
stack1.pop();
stack1.push(tokens[i]);
}

行 getPrecedence(stack1.top()> getPrecedence(tokens[I]) 是问题所在。特别是，在 stack1.top() 上运行 getPrecedence。此函数基本上接收一个字符串并将其与保存所有存储运算符的向量进行比较。当它找到索引时，它会返回该索引的优先级(它们按顺序设置所有索引)。我不明白为什么我不能以这种方式调用这个函数。stack1.top() 将只给出一个将传递和比较的字符串。有什么想法吗？

Answer 1

想通了。有一些事情正在发生，但困扰程序的主要问题是我在不应该的时候从堆栈中弹出东西，导致堆栈是空的，所以它永远不会进入

while (getPrecedence(stack1.top()) > getPrecedence(tokens[i]) || (getPrecedence(stack1.top()) == getPrecedence(tokens[i]) &&
getAssociativity(tokens[i]) == 'L') && stack1.top() != "(")