问题
I've written a few methods for a calculator. One, which evaluates an entered Postfix expression and another, which transfers an entered infix expression into a postfix expression.
Both these methods allow multi digit integers aswell as floats for the number input types.
Now for my question:
I want to include the negative input in both these methods e.g. Infix: "3 * (-1)".
However I'm kinda lacking an idea on how to implement this problem. Maybe someone can give me ideas or code examples.
I'm including both methods below. A few simple methods are used in them which aren't shown here, but most of the function names should explain them very well. Due to this fact I left them out to keep things as short as possible.
string InfixToPostfix(string expression)
{
string postfix = "";
stack <char> S;
for (int i = 0; i < expression.length(); i++)
{
if (expression[i] == ' ') continue; // Falls leerzeichen oder ',' gefunden mit nächster Iteration weiter machen
else if (IsOperator(expression[i])) // Falls ein Operator gefunden wurde:
{
while (!S.empty() && S.top() != '(' && HasHigherPrecedence(S.top(), expression[i])) // Wenn Operator auf Stack höhere Precedence hat, dann diesen an String anhängen und vom Stack nehmen
{
postfix += S.top();
postfix += ' ';
S.pop();
}
S.push(expression[i]); // Wenn Operator die Bedingungen in der while Schleife nicht erfüllt diesen auf Stack legen
}
else if (isDigit(expression[i]) || isComma(expression[i])) //Wenn ein Digit gefunden wurde diesen an String anhängen
{
postfix += expression[i];
if (i+1 >= expression.length() || (!isDigit(expression[i + 1]) && !isComma(expression[i+1]))) //Wenn die nächste Zahl kein Digit ist, dann leerzeichne an String anhängen
{
postfix += ' ';
}
}
else if (expression[i] == '(') // '(' wird auf Stack gepusht
{
S.push(expression[i]);
}
else if (expression[i] == ')') // Wenn ')' gefunden wird, dann:
{
while (!S.empty() && S.top() != '(') // Sofern Stack nicht leer und das oberste Element des Stacks nicht eine Klammer auf ist wird das oberste Element des Stacks dem String angehängt
{
postfix += S.top();
S.pop();
}
S.pop(); //ansonsten wird '(' einfach vom Stack genommen
}
}
while (!S.empty()) // Am Ende der Expression werden alle verbleibenden Elemente vom Stack genommen und Leerzeichen eingefügt
{
postfix += S.top();
postfix += ' ';
S.pop();
}
return postfix; // Rückgabe der jetzt in Postfix vorliegenden Expression
}
//Löst eine Aufgabe in Postfix Notation
float EvaluatePostfix(string expression)
{
stack<float> S;
float j;
for (int i = 0; i< expression.length(); i++) {
if (expression[i] == ' ') continue; // wenn leer oder ',' mit nächster iteration weiter machen
else if (IsOperator(expression[i])) { //Wenn Operator nehme die Operanden vom Stack und wende den Operator an
float operand2 = S.top();
S.pop();
float operand1 = S.top();
S.pop();
float result = PerformOperation(expression[i], operand1, operand2);
S.push(result); //Das Ergebnis zurück auf Stack legen
}
else if (isDigit(expression[i]))
{
float operand = 0;
while (i<expression.length() && isDigit(expression[i]))
{
operand = (operand * 10) + (expression[i] - '0'); //wenn rechts einer Zahl eine weitere Zahl steht, kann operand mal 10 genommen werden und die rechts stehende zahl addiert werden
i++;
}
if (i < expression.length() && isComma(expression[i]))
{
j = 1.0;
i++;
while (i < expression.length() && isDigit(expression[i]))
{
operand = operand + ((expression[i] - '0') / pow(10.0, j));
i++;
j++;
}
}
i--; //Verhindert einen Skip des Operators, i wird sowohl in der while schleife als auch im for hochgezählt
S.push(operand); //Der Operand wird auf den Stack gepusht
}
}
return S.top(); //Stack sollte ein element besitzen, somit ist dies das Ergebnis
}
回答1:
I don't have a full solution for you, but here are a couple of tips.
I suggest inserting an abstraction layer that reads characters and produces tokens before trying to understand the order of operations. The expression "(42 + 1) - -3" would then become the list { '(', 42, '+', 1, ')', '-', '-', 3 }. A token is often implemented as a class with a type enum (e.g., OPERATOR or NUMBER) and a value (e.g., either a char or a float). (Advanced: You could then convert your tokens into an expression tree, but that might not be necessary here.) This is a little more work, but it makes things much easier to understand than direct string-parsing.
Once you've done that, the key is to determine whether the '-' symbol is intended as infix subtraction or as prefix negation. To do that, look at the previous token (if any). If it's an expression-terminator such as ')' or a number, it's infix subtraction. If there is no such token, or it is some other token, it's prefix negation.
回答2:
This is what I did:
class expression
{
//constructors and destructors
public:
expression();
~expression();
//member properties
private:
std::string expr;
std::string postfixExpr;
//member methods
public:
// accepts the expression from the user
void input();
//prints the accepted expression
void output();
//evaluates the accepted expression
float eval();
//converts infix expression to postfix
void convertToPostfix();
friend bool isOperator(char);
};
inline bool isOperator(char c)
{
if (c == '+' || c == '-' || c == '*' || c == '/' || c == '(' || c == ')' || c == '#')
return true;
return false;
}
void expression::input()
{
std::cin >> this->expr;
}
float expression::eval()
{
std::stack<float> s;
float op1, op2;
for (int i = 0; i < expr.length(); i++)
{
if (expr[i] == '(')
{
float sum = 0;
bool flag = false;
while (expr[++i] != ')')
{
if (expr[i] == '-') {
flag = true;
i++;
}
sum = sum * 10.0 + (float(expr[i]) - float('0'));
}
if (flag)
sum = -sum;
s.push(sum);
continue;
}
else if (!isOperator(expr[i]))
{
s.push(float(expr[i]) - float('0'));
}
else
{
op2 = s.top();
s.pop();
op1 = s.top();
s.pop();
switch (expr[i])
{
case '+':
s.push(op1 + op2);
break;
case '-':
s.push(op1 - op2);
break;
case'*':
s.push(op1*op2);
break;
case '/':
s.push(op1 / op2);
break;
default:
std::cerr << "Wrong operator" << std::endl;
return 0;
}
}
}
return s.top();
}
In my implementation, the only hitch is to use '(' and ')' as delimiters. Secondly, I also made std::string expr as a property in my expression class. This works fine. But mind well, do input your expression in postfix notation as I still have to include the convertToPostfix function. But I guess, you will figure that out on your own :)
Happy Coding :)
来源：https://stackoverflow.com/questions/40553697/how-can-i-accept-negative-values-in-postfix-and-infix-notation