我正在做一些与此用户非常相似的事情： Error: deque iterator not dereferenceable
我一直在寻找一个年龄，但我无法看到我出错的地方。另一张海报的解决方案是找到一个地方，他试图从一个零元素的双端队列弹出或顶部。我无法在我的代码中找到我在做的地方。
编辑：我怀疑问题出在SYAlg或OSProcess中，如果有帮助的话。
// TestCalculator.cpp : main project file.
#include <string>
#include <iostream>
#include <locale>
#include <ctype.h>
#include <vector>
#include <deque>
using namespace System;
using namespace std;
//using std::string;
bool lastCharDigit = true;
string RawString; //Contains the raw equation the user types in.
deque<string> TokenEquation(1); //Contains the equation in tokenised infix form.
deque<string> RPNEquation; //Contains the equation in tokenised RPN form.
deque<string> OperatorStack; //Used as part of the Shunting Yard Algorithm
deque<string> SolverStack; //Used to solve the RPN Equation.
locale loc; //Used to verify digits.
//START FUNCTION DECLARATION
int main();
void tokeniser(string RawEquation);
void SYAlg();
void OSProcess(string newOperator);
void Solver();
//END FUNCTION DECLARATION
int main()
{
cout << "Please enter a valid infix notation equation, without parenthesis.\n";
cin >> RawString;
tokeniser(RawString);
cout << "\n";
system("pause");
return 0;
}
void tokeniser(string RawEquation)
{
int testCharPos = -1; // Initialise the index of the raw string
int tokenVectorPos = 0; // Initialise the token array position
int tokenVectorPrintPos = 0; // Initialise the print position
for (int eLength = RawEquation.length(); eLength != 0; eLength--) // For each character in the Raw string...
{
testCharPos++; // Increment the char we're testing
char testChar = RawEquation.at(testCharPos); // Establish the current test char
if (isdigit(testChar, loc)) //If the testchar is a digit
{
if (lastCharDigit) //If the last character was a digit
{
TokenEquation[tokenVectorPos] += testChar; //Append the tested char to the current token array pos
}
if (!lastCharDigit) //If the last character was not a digit
{
TokenEquation.push_back(string(1, testChar)); //Establish a new element with the testchar in it.
tokenVectorPos++;
}
lastCharDigit = true;
}
if (!isdigit(testChar, loc))//If the testchar is not a digit
{
TokenEquation.push_back(string(1, testChar)); //Establish a new element with the testchar in it.
tokenVectorPos++;
lastCharDigit = false;
}
}
cout << "The tokens of that equation are:\n\n"; //Outputs the tokens for testing purposes.
for (int tokenLength = TokenEquation.size(); tokenLength != 0; tokenLength--)
{
cout << " " << TokenEquation[tokenVectorPrintPos];
cout << "\n";
tokenVectorPrintPos++;
}
SYAlg(); //Call the SYAlg.
}
void SYAlg() //This function uses Shunting Yard Algorithm to convert the Infix tokens to RPN.
{
cout << TokenEquation.size();
for (int testtokenLength = TokenEquation.size(); testtokenLength != 0; testtokenLength--) //For each token in the tokenised deque
{
if (isdigit(TokenEquation.front().at(0), loc)) //Check if it's a number
{
RPNEquation.push_back(TokenEquation.front()); //Add the first raw token to the RPN Equation
TokenEquation.pop_front(); //Pop the token from the deque
}
if (!isdigit(TokenEquation.front().at(0), loc)) //If it's an operator
{
OSProcess(TokenEquation.front()); //Run the SYAlg operator stack procedure. NB This will pop the front of the TokenEquation for you.
}
}
cout << "The tokens of that equation are:\n\n"; //Outputs the tokens for testing purposes.
int RPNPrintPos = 0;
for (int tokenLength = RPNEquation.size(); tokenLength != 0; tokenLength--)
{
cout << " " << RPNEquation[RPNPrintPos];
cout << "\n";
RPNPrintPos++;
}
}
void OSProcess(string newOperator) //This function processes the Operator Stack
{
bool PushedNewOperator = false;
std::string newOpSTD = newOperator; //Creates an std::string version of the argument for easier comparison.
while (PushedNewOperator == false){ //As long as the new operator is still waiting to go to the stack
if (!OperatorStack.empty()) //If there's already an operator on the stack
{
if (newOpSTD == "/" || "*")
{
std::string OSBackSTD = OperatorStack.back(); //Create an STD version of the back of the OpStack for comparison.
if (OSBackSTD == "+" || "-")
{
OperatorStack.push_back(newOperator); //Add the tested operator to the stack
TokenEquation.pop_front(); //And pop it from the token equation
PushedNewOperator = true; //Set the flag variable to true so we stop looping
}
else
{
RPNEquation.push_back(OperatorStack.back()); //Add the top of the operator stack to the equation
OperatorStack.pop_back(); //Pop this back
}
}
else
{
RPNEquation.push_back(OperatorStack.back()); //Add the top of the operator stack to the equation
OperatorStack.pop_back(); //Pop this back
}
}
if (OperatorStack.empty())
{
OperatorStack.push_back(newOperator); //Add the tested operator to the stack
TokenEquation.pop_front(); //And pop it from the token equation
PushedNewOperator = true; //Set the flag variable to true so we stop looping
}
}
//For each operator on the stack, until the following statement returns false...
//Check if the precedence of newOperator is less than or equal to the top operator.
}
void Solver() //This function solves the RPNEquation
{
//Push each token to the solver stack
//If you push an operator, solve it against the stack
//When the RPN equation is empty and the solver stack only has one token in it, you have a solution
}
答案 0 :(得分：0)
一个主要问题是多行说if (newOpSTD == "/" || "*")，或者说有什么效果。这些需要更改为if (newOpSTD.compare("/") == 0 || newOpSTD.compare("*") == 0)。
我认为这些检查失败意味着while循环将它们转变为while(true)。