我正在使用 Jesse Brown 的调车场算法实现的修改版本。我正在尝试修改变量系统以基本上执行符号数学，而不是为变量分配双精度值。例如，我不想简单地声明 pi = 3.14，而是希望它只在解决方案中包含 pi。所以 1+2+pi 将导致 3+pi。
代码如下。我刚刚开始弄乱它，并没有做很多。有没有人有任何想法？
// Source: http://www.daniweb.com/software-development/cpp/code/427500/calculator-using-shunting-yard-algorithm#
// Author: Jesse Brown
// Modifications: Brandon Amos
#include <cstdlib>
#include <iostream>
#include <sstream>
#include <stdexcept>
#include "shunting_yard.h"
using namespace std;
#define isvariablechar(c) (isalpha(c) || c == '_')
TokenQueue_t calculator::toRPN(const char* expr,
map<string, string>* vars,
map<string, int> opPrecedence) {
TokenQueue_t rpnQueue; stack<string> operatorStack;
bool lastTokenWasOp = true;
// In one pass, ignore whitespace and parse the expression into RPN
// using Dijkstra's Shunting-yard algorithm.
while (*expr && isspace(*expr)) ++expr;
while (*expr) {
if (isdigit(*expr)) {
// If the token is a number, add it to the output queue.
char* nextChar = 0;
double digit = strtod(expr, &nextChar);
# ifdef DEBUG
cout << digit << endl;
# endif
rpnQueue.push(new Token<double>(digit));
expr = nextChar;
lastTokenWasOp = false;
}
else if (isvariablechar(*expr)) {
// If the function is a variable, resolve it and
// add the parsed number to the output queue.
if (!vars) {
//throw domain_error(
//"Detected variable, but the variable map is null.");
}
stringstream ss;
ss << *expr;
++expr;
while (isvariablechar(*expr)) {
ss << *expr;
++expr;
}
string key = ss.str();
map<string, string>::iterator it = vars->find(key);
if (it == vars->end()) {
//throw domain_error(
//"Unable to find the variable '" + key + "'.");
}
string val = vars->find(key)->second;
# ifdef DEBUG
cout << val << endl;
# endif
rpnQueue.push(new Token<string>(val));;
lastTokenWasOp = false;
}
else {
// Otherwise, the variable is an operator or paranthesis.
switch (*expr) {
case '(':
operatorStack.push("(");
++expr;
break;
case ')':
while (operatorStack.top().compare("(")) {
rpnQueue.push(new Token<string>(operatorStack.top()));
operatorStack.pop();
}
operatorStack.pop();
++expr;
break;
default:
{
// The token is an operator.
//
// Let p(o) denote the precedence of an operator o.
//
// If the token is an operator, o1, then
// While there is an operator token, o2, at the top
// and p(o1) <= p(o2), then
// pop o2 off the stack onto the output queue.
// Push o1 on the stack.
stringstream ss;
ss << *expr;
++expr;
while (*expr && !isspace(*expr) && !isdigit(*expr)
&& !isvariablechar(*expr) && *expr != '(' && *expr != ')') {
ss << *expr;
++expr;
}
ss.clear();
string str;
ss >> str;
# ifdef DEBUG
cout << str << endl;
# endif
if (lastTokenWasOp) {
// Convert unary operators to binary in the RPN.
if (!str.compare("-") || !str.compare("+")) {
rpnQueue.push(new Token<double>(0));
}
else {
//throw domain_error(
//"Unrecognized unary operator: '" + str + "'.");
}
}
while (!operatorStack.empty() &&
opPrecedence[str] <= opPrecedence[operatorStack.top()]) {
rpnQueue.push(new Token<string>(operatorStack.top()));
operatorStack.pop();
}
operatorStack.push(str);
lastTokenWasOp = true;
}
}
}
while (*expr && isspace(*expr)) ++expr;
}
while (!operatorStack.empty()) {
rpnQueue.push(new Token<string>(operatorStack.top()));
operatorStack.pop();
}
return rpnQueue;
}
double calculator::calculate(const char* expr,
map<string, string>* vars) {
// 1. Create the operator precedence map.
map<string, int> opPrecedence;
opPrecedence["("] = -1;
opPrecedence["<<"] = 1; opPrecedence[">>"] = 1;
opPrecedence["+"] = 2; opPrecedence["-"] = 2;
opPrecedence["*"] = 3; opPrecedence["/"] = 3;
// 2. Convert to RPN with Dijkstra's Shunting-yard algorithm.
TokenQueue_t rpn = toRPN(expr, vars, opPrecedence);
// 3. Evaluate the expression in RPN form.
stack<double> evaluation;
while (!rpn.empty()) {
TokenBase* base = rpn.front();
rpn.pop();
Token<string>* strTok = dynamic_cast<Token<string>*>(base);
Token<double>* doubleTok = dynamic_cast<Token<double>*>(base);
if (strTok) {
string str = strTok->val;
/*if (evaluation.size() < 2) {
throw domain_error("Invalid equation.");
}*/
if (str.compare("pi")){
cout << "its pi!" << endl;
}
double right = evaluation.top(); evaluation.pop();
double left = evaluation.top(); evaluation.pop();
if (!str.compare("+")) {
evaluation.push(left + right);
}
else if (!str.compare("*")) {
evaluation.push(left * right);
}
else if (!str.compare("-")) {
evaluation.push(left - right);
}
else if (!str.compare("/")) {
evaluation.push(left / right);
}
else if (!str.compare("<<")) {
evaluation.push((int)left << (int)right);
}
else if (!str.compare(">>")) {
evaluation.push((int)left >> (int)right);
}
else {
cout << "Unknown Operator" << endl;
//throw domain_error("Unknown operator: '" + str + "'.");
}
}
else if (doubleTok) {
evaluation.push(doubleTok->val);
}
else {
//throw domain_error("Invalid token.");
}
delete base;
}
return evaluation.top();
}