我正在努力实现Shunting Yard Algorithm以评估简单表达式。该代码似乎可以正常工作,但是如果有空格,则会崩溃。这是令人惊讶的,因为对空白进行了专门的检查,似乎根本无法捕获它。
using namespace std;
mpz_class exprToTokens(string expression);
int precedence(const char op);
mpz_class applyOperation(const mpz_class a, const mpz_class b, const char op);
int main() {
try {
while(true)
{
cout << "enter an expression: ";
string expr;
cin >> expr;
if( expr == "e")
{
break;
}
cout << "output: " << exprToTokens(expr) << endl;
}
} catch( const std::exception & ex ) {
cerr << "message: " << ex.what() << endl;
}
return 0;
}
mpz_class exprToTokens(string expression)
{
stack<char> operators;
stack<mpz_class> output;
unsigned int i = 0;
while(i < expression.length())
{
if(isspace(static_cast<unsigned char>(expression.at(i))))//skip white space
{
cout << "is space" << endl;//never happens
i++;
continue;
}
else if(isdigit(expression[i]))
{
unsigned int j = i;
while(i < expression.length() && isdigit(expression[j]))
{
j++;
}
const string number = expression.substr(i, j-i);
const mpz_class term(number);
output.push(term);
i = j;
continue;
}
else//token is an operator
{
while(!operators.empty() && precedence(operators.top() >= precedence(expression[i])))
{
const mpz_class val1 = output.top();
output.pop();
const mpz_class val2 = output.top();
output.pop();
const char op = operators.top();
operators.pop();
output.push(applyOperation(val1, val2, op));
}
operators.push(expression[i]);
}
i++;
}
/*process remaining operations and values on stacks*/
while(!operators.empty())
{
const mpz_class val2 = output.top();//something bad happens here when spaces are around operator
output.pop();
const mpz_class val1 = output.top();
output.pop();
const char op = operators.top();
operators.pop();
output.push(applyOperation(val1, val2, op));
}
return output.top();
}
int precedence(const char op)
{
if(op == '+' || op == '-')
return 1;
if(op == '*' || op == '/')
return 2;
return 3;
}
mpz_class applyOperation(const mpz_class a, const mpz_class b, const char op)
{
switch(op)
{
case '+': return a + b;
case '-': return a - b;
case '*': return a * b;
case '/': return a / b;
default: throw invalid_argument("syntax not recognized");
}
}
例如,
3+3
给出6
结果,但3 + 3
导致分段错误。有任何想法吗?排除问题:Shunting Yard算法将中缀转换为后缀表示法。因此严格来说,修改算法以实际评估表达式是很常见的,但是分流场算法仍然存在吗?对于普通算法,是否还需要另一种算法来仍然以后缀表示法评估表达式?
最佳答案
cin >> expr;
在空格处分割输入,因此将读取三个单独的表达式:3
,+
和3
。要读取整个表达式,请使用 std::getline()
。
您的程序可能无法解析第一个,因为此循环不检查j
:
while(i < expression.length() && isdigit(expression[j]))
{
j++;
}