北林OJ基于栈的后缀表达式求值

思路1：（较易理解）

#include<iostream>
#include <iomanip>
using namespace std;
#define OK 1
#define ERROR 0
#define MaxSize 100
typedef double ElemType;
//定义栈_顺序栈

struct Stack
{
	ElemType* top;
	ElemType* base;
	int stacksize;
};
int IsFull(Stack s);
int IsEmpty(Stack s);
//初始化
int InitStack(Stack& s)
{
	s.base = new ElemType[MaxSize];
	if (!s.base)
		return ERROR;
		s.top = s.base;
	s.stacksize = MaxSize;
	return OK;
}
//入栈
int PushbackStack(Stack& s, ElemType e)
{
	//判断栈是否满
	if (IsFull(s))
		return ERROR;
	*(s.top) = e;//error
	(s.top)++;
	return OK;
}
//出栈
ElemType Popback(Stack& s)
{
	//判断栈是否空
	if (IsEmpty(s))
		return ERROR;
	s.top--;
	return *(s.top);
}
//判满
int IsFull(Stack s)
{
	if (s.top - s.base == s.stacksize)
		return 1;
	return 0;
}
//判空
int IsEmpty(Stack s)
{
	if (s.base == s.top)
		return 1;
	return 0;
}
//获取栈顶元素
ElemType StackTop(Stack s)
{
	//判断栈是否空
	if (IsEmpty(s))
		return ERROR;
	return *(s.top - 1);
}
//清除栈
int ClearStack(Stack& s)
{
	s.top = s.base;
	return OK;
}

int Destroy(Stack& s)
{
	delete s.base;
	s.top = s.base = NULL;
	return OK;
}
//计算
double calculating(double num1, char s, double num2)
{
	switch (s)
	{
	case'+':
		return (double)(num1 + num2)*(1.0);
	case'-':
		return (double)(num1 - num2) * (1.0);
	case'*':
		return (double)(num1 * num2) * (1.0);
	case'/':
		return (double)((num1*1.00) /(double) num2) * (1.0);
	}
}
double EvaluateExpressionSuffix(int &flag)
{
	//OPND:存放数字
	Stack OPND;
	char ch;
	
	InitStack(OPND);
	cin >> ch;
	if (ch == '=')
	{
		flag = 0;
		return ERROR;
	}
	do
	{
		if (isdigit(ch))
		{
			PushbackStack(OPND, (double)(ch - 48));//将字符转为数字，如果是ch,结果为-1
			cin >> ch;
		}
		else
		{
			
			PushbackStack(OPND, calculating(Popback(OPND),
				ch, Popback(OPND)));//计算结果并入栈
			cin >> ch;
		}
	} while (ch != '=');
	return Popback(OPND);
	
}
void test1()
{
	int flag = 1;
	while (1)
	{
		double outcome=EvaluateExpressionSuffix(flag);
		if (!flag)
			return;
		cout << fixed << setprecision(2) << outcome << endl;
	}
}
int main()
{

	test1();
	
	return 0;
}

遇到的问题：类型的转换：从char类型转换到其他类型

double类型数据精度：保留两位小数

思路二：将每行输入存放到数组里，从数组中读取

栈的操作与之前相同，下面便不再写了

//计算
double calculating(double num1, char s, double num2)
{
	switch (s)
	{
	case'+':
		return (double)(num1 + num2)*(1.0);
	case'-':
		return (double)(num1 - num2) * (1.0);
	case'*':
		return (double)(num1 * num2) * (1.0);
	case'/':
		return (double)((num1*1.00) /(double) num2) * (1.0);
	}
}
Status EvaluateExpressionSuffix(char *s)
{
	//OPND:存放数字
	Stack OPND;
	InitStack(OPND);
	int i = 0;
	if (s[i] == '=')//防止死循环
		return ERROR;
	while (s[i] != '='&&s[i]!='\0')
	{
		if (isdigit(s[i]))
		{
			PushbackStack(OPND, (double)(s[i] - 48));//将字符转为数字，如果是ch,结果为-1

		}
		else if (s[i] == ' ')
			;
		else
		{
			//error
			PushbackStack(OPND, calculating(Popback(OPND),
				s[i], Popback(OPND)));//计算结果并入栈
		}
		i++;
	} 
	cout << fixed << setprecision(2) << Popback(OPND) << endl;
	return OK;
}
void test1()
{
	while (1)
	{
		char s[MaxSize];//死循环打印0.00，为什么？？
		//因为cin.get操作后会在缓冲区存留0
		cin.get(s, MaxSize);//读入数据并存放到数组s中,在字符串尾部加'\0'
		char ch = cin.get();//	清除缓冲区结尾残留的字符\0
		int flag = EvaluateExpressionSuffix(s);
		if (!flag )
			break;
	}
	
	return;
}
int main()
{

	test1();
	
	return 0;
}

问题：计算1 2/=时，结果总是为0.00，

解决：将整数除法转化为小数除法，影响转换成double在计算，注意顺序

使用cin>>读取每一行时遇到空格会停止，所以要使用cin.get函数

对cin函数不够了解导致出错

解决：查找资料，并自己写一些简单的例子测试，多多调试加深理解

cin相关知识点：https://blog.csdn.net/a3192048/article/details/80303547

如果对cin及其函数不太了解，推荐第一种写法