表达式求值

// exam1.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <iostream>
#include <stack>
using namespace std;

int priority(char op)
{
	switch(op)
	{
	case '(':
		return 1;
	case '+':
	case '-':
		return 2;
	case '*':
	case '/':
		return 3;
	case ')':
		return 4;
	case '#':
		return 5;
	}
}

bool cmp_lg(char op1,char op2)
{
	int proi1=priority(op1);
	int proi2=priority(op2);

	if(proi1>=proi2)
	{
		return true;
	}
	else
	{
		return false;
	}
}

int cal_func(char op,int num1,int num2)
{
	switch(op)
	{
	case '+':
		return num1+num2;
	case '-':
		return num1-num2;
	case '*':
		return num1*num2;
	case '/':
		return num1/num2;
	default:
		return num1+num2;
	}
}

void cal(void)
{
	cout<<"Please enter the formula whose end is '#'..."<<endl;
	char ch;
	stack<int> num;
	stack<char> op;

	while(1)
	{
		cin>>ch;
		if(ch=='#')
		{
			break;
		}

		if(ch>='0' && ch<='9')
		{
			num.push(ch-'0');
		}
		else
		{
			if(op.empty())
			{
				op.push(ch);
			}
			else
			{
				char op1,op2;

				op1=op.top();
				op2=ch;

				if(ch=='(')
				{
					op.push(ch);
				}
				else if(cmp_lg(op1,op2))
				{
					int result,num1,num2;
					num1=num.top();
					num.pop();
					num2=num.top();
					num.pop();

					result=cal_func(op1,num1,num2);
					num.push(result);

					op.pop();
					op.push(op2);
				}
				else
				{
					if(ch==')')
					{
						int result;
						int num1,num2;
						char op0;

						num1=num.top();
						num.pop();
						num2=num.top();
						num.pop();

						op0=op.top();
						op.pop();
						op.pop();
						result=cal_func(op0,num1,num2);
						num.push(result);
					}
					else
					{
						op.push(ch);
					}
				}
			}
		}
	}
	
	while(!num.empty() && !op.empty())
	{
		int result;
		int num1,num2;
		char op0;

		num1=num.top();
		num.pop();
		num2=num.top();
		num.pop();

		op0=op.top();
		op.pop();
		result=cal_func(op0,num1,num2);
		num.push(result);
	}

	cout<<"The result is "<<num.top()<<"."<<endl;
}

int main(void)
{
	cal();

	system("pause");
	return 0;
}

表达式求值,布布扣,bubuko.com

表达式求值

原文：http://blog.csdn.net/cjc211322/article/details/38356065