我正在尝试编写一个代码，该代码将采用格式类似于
的表达式(((p∨q)∧((q→r)⊕(p∧r)))↔(r∧q))→(p∨r)
它需要打印出类似
的真值表p q r (p V q) (q→r) etc... until it gets to (((p∨q)∧((q→r)⊕(p∧r)))↔(r∧q))→(p∨r)
t t t t t
t t f t f
t f t t t
t f f t t
f t t t t
f t f t f
f f t f t
f f f f t
我设计了一种处理XOR并隐含运算符的方法，但我意识到只有当运算符位于内括号内时它才有效，而不是当运算符位于两组括号之间时...
我已经花了太多时间研究这个问题，有人可以让我深入了解我可以使用的东西吗？
现在我想更多了，我可以尝试在第一个（）中读取，然后将第二个（）等作为单个变量读取，将它们转换为计算机将理解的布尔值，然后将它们注入按照正确的顺序进入正确的格式？
我正在处理的源代码是
#include <iostream>
#include <iomanip>
#include <fstream>
#include <string>
#include <vector>
#include <algorithm>
int main()
{
bool p[8] = { true, true, true, true, false, false, false, false }; // Declare the init T/F values
bool q[8] = { true, true, false, false, true, true, false, false };
bool r[8] = { true, false, true, false, true, false, true, false };
std::string expression;
std::cout << "Enter expression (using ^ for AND, V for OR, X for XOR, and I for Implies (-->): \n";
std::getline(std::cin, expression);
int i = 0;
int j = 0;
std::vector<std::string> exprVector;
for (std::string::iterator it = expression.begin; it != expression.end(); it++)
{
if (expression.at(i) == 'p')
{
exprVector[i] = "p[i]";
}
else if (expression.at(i) == 'q')
{
exprVector[i] = "q[i]";
}
else if (expression.at(i) == 'r')
{
exprVector[i] = "r[i]";
}
else if (expression.at(i) == '(')
{
exprVector[i] = "(";
}
else if (expression.at(i) == ')')
{
exprVector[i] = ") ";
}
else if (expression.at(i) == '^')
{
exprVector[i] = "&&";
}
else if (expression.at(i) == 'V')
{
exprVector[i] = "||";
}
else if (expression.at(i) == 'X')
{
char temp1;
char temp2;
i--;
if (exprVector[i] == "p[i]") // (p XOR q)
{
i++;
i++;
if (exprVector[i] == "q[i]")
{
i--;
//clear vector[i] from exprVector
i--;
//clear vector[i] from exprVector
exprVector[i] = "((p[i] || q[i]) && ((p[i] && q[i]) == false))";
}
else if (exprVector[i] == "r") // (p XOR r)
{
i--;
//clear vector[i] from exprVector
i--;
//clear vector[i] from exprVector
exprVector[i] = "((p[i] || r[i]) && ((p[i] && r[i]) == false))";
}
}
else if (exprVector[i] == "q") // (q XOR p)
{
i++;
i++;
if (exprVector[i] == "p")
{
i--;
//clear vector[i] from exprVector
i--;
//clear vector[i] from exprVector
exprVector[i] = "((q[i] || p[i]) && ((q[i] && p[i]) == false))";
}
else if (exprVector[i] == "r") // (q XOR r)
{
i--;
//clear vector[i] from exprVector
i--;
//clear vector[i] from exprVector
exprVector[i] = "((q[i] || r[i]) && ((q[i] && r[i]) == false))";
}
}
else if (exprVector[i] == "r")
{
i++;
i++;
if (exprVector[i] == "p") // (r XOR p)
{
i--;
//clear vector[i] from exprVector
i--;
//clear vector[i] from exprVector
exprVector[i] = "((r[i] || p[i]) && ((r[i] && p[i]) == false)";
}
if (exprVector[i] == "q") // (r XOR q)
{
i--;
//clear vector[i] from exprVector
i--;
//clear vector[i] from exprVector
exprVector[i] = "((r[i] || q[i]) && ((r[i] && q[i]) == false)";
}
}
}
else if (expression.at(i) == 'I')
{
if (exprVector[i] == "p[i]") // (p ---> q)
{
i++;
i++;
if (exprVector[i] == "q[i]")
{
i--;
//clear vector[i] from exprVector
i--;
//clear vector[i] from exprVector
exprVector[i] = "((p[i] == true) || (q[i] == false)";
}
else if (exprVector[i] == "r") // (p ---> r)
{
i--;
//clear vector[i] from exprVector
i--;
//clear vector[i] from exprVector
exprVector[i] = "((p[i] == true) || (r[i] == false)";
}
}
else if (exprVector[i] == "q") // (q ---> p)
{
i++;
i++;
if (exprVector[i] == "p")
{
i--;
//clear vector[i] from exprVector
i--;
//clear vector[i] from exprVector
exprVector[i] = "((q[i] == true) || (p[i] == false))";
}
else if (exprVector[i] == "r") // (q ---> r)
{
i--;
//clear vector[i] from exprVector
i--;
//clear vector[i] from exprVector
exprVector[i] = "((q[i] == true) || (r[i] == false))";
}
}
else if (exprVector[i] == "r")
{
i++;
i++;
if (exprVector[i] == "p") // (r ---> p)
{
i--;
//clear vector[i] from exprVector
i--;
//clear vector[i] from exprVector
exprVector[i] = "((r[i] == true) || (p[i] == false))";
}
if (exprVector[i] == "q") // (r ---> q)
{
i--;
//clear vector[i] from exprVector
i--;
//clear vector[i] from exprVector
exprVector[i] = "((r[i] == true) || (q[i] == false))";
}
}
else if (expression.at(i) == '!')
{
i++
if (exprVector[i] == "p")
{
//clear vector[i] from exprVector
i--;
//clear vector[i] from exprVector
exprVector[i] == "(p[i] == false)"
}
if (exprVector[i] == "q")
{
//clear vector[i] from exprVector
i--;
//clear vector[i] from exprVector
exprVector[i] == "(p[i] == false)"
}
if (exprVector[i] == "r")
{
//clear vector[i] from exprVector
i--;
//clear vector[i] from exprVector
exprVector[i] == "(p[i] == false)"
}
}
else if (expression.at(i) == ' ')
{
}
i++;
j++;
}
}
答案 0 :(得分：0)
您是否考虑过递归下降解析器？ http://en.wikipedia.org/wiki/Recursive_descent_parser
可能还有其他一些解析实现。然后，您可以生成一个表示表达式的树，然后您可以为所有可能的输入计算树。