初级数据结构--栈在表达式求职中的应用2

只支持正确格式表达式，判断非法表达式逻辑没写太多

纯个人理解，指套入了部分表达式测试，如有错误欢迎指出

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <ctype.h>
typedef struct Operand
{
	int and[MAX_SIZE];
	int top;
}Operand;
typedef struct Operator
{
	int top;
	char ator[MAX_SIZE];
}Operator;

//初始化两个栈
void Init_Oper(Operand *p1, Operator *p2)
{
	int i = 0;
	for (i = 0; i < MAX_SIZE; i++)
	{
		p1->and[i] = 0;
		p2->ator[i] = 0;
	}
	p1->top = 0;
	p2->top = 0;
}
//计算
bool Ifator(char c)
{
	switch (c)
	{
	case '+':
	case '-':
	case '*':
	case '/':
	case '(':
	case ')':
		return  true;
	default:
		return false;
	}
}
int Calculate(char ator,int x,int y)
{
	switch (ator)
	{
	case '+':
		return x + y;
		break;
	case '-':
		return x - y;
		break;
	case '*':
		return x*y;
		break;
	case '/':
		return x / y;
		break;
	default:
		break;
	}
}
void Out(Operand *p1, Operator *p2)
{
	int ret = 0;
	ret = Calculate(p2->ator[p2->top - 1],
		p1->and[p1->top - 2],
		p1->and[p1->top - 1]);
	p1->and[p1->top - 2] = ret;
	p1->and[p1->top - 1] = 0;
	p1->top--;
	p2->ator[p2->top - 1] = 0;
	p2->top--;
}
bool GetAnswer(Operand *p1, Operator *p2, char *expp)
{
	bool bracket = false;
	while (*expp || (p2->top&&p1->top!=1))
	{
		int n = 0;
		if (isdigit(*expp))//压入操作数
		{
			while (*expp&&isdigit(*expp))
			{
				n = n * 10 + *expp-'0';//char型数字需要循环记录完整操作数
				expp++;
			}
			p1->and[p1->top] = n;
			p1->top++;
		}
		else if (Ifator(*expp))//压入操作符 || *expp == '(' || *expp == ')'
		{
			if (bracket)
			{
				p2->ator[p2->top] = *expp;
				p2->top++;
				expp++;
				bracket = false;
			}

			if (0 == p2->top&&*expp!=')')
			{
				p2->ator[p2->top] = *expp;
				p2->top++;
				expp++;
			}
			else
			{
				if (*expp == '('||p2->ator[p2->top-1]=='(')
				{
					p2->ator[p2->top] = *expp;
					p2->top++;
					expp++;
					if(*expp == '(')
						bracket=true;
				}
				if (*expp == ')')
				{
					while (p2->ator[p2->top - 1] != '(')
					{
						if (!(p2->top))
						{
							Init_Oper(p1, p2);//非法表达式返回false并清空两个栈
							return false;
						}
						Out(p1, p2);
					}
					p2->ator[p2->top-1] = 0;
					p2->top--;
					expp++;
				}
				if (*expp == '+' || *expp == '-')//下一个运算符为+ - 出栈
				{
					Out(p1, p2);
				}
				else if (*expp == '*' || *expp == '/')
				{
					//+ - 压栈
					if (p2->ator[p2->top - 1] == '+' || p2->ator[p2->top - 1] == '-')
					{
						p2->ator[p2->top] = *expp;
						p2->top++;
						expp++;
					}
					// * / 出栈
					else if (p2->ator[p2->top - 1] == '*' || p2->ator[p2->top - 1] == '/')
					{
						Out(p1, p2);
					}
				}
			}
			//expp++;
		}
		else if (!(*expp)&&p2->top)
		{
			Out(p1, p2);
		}
		else
			return false;
	}
	if (1==(p1->top - 1) && !(p2->top - 1))
		return true;
	else
		return false;
}
int main()
{
	Operand oand;
	Operator oator;
	char exp[] = "(1+2*1)*3+2*2*2";//17
	Init_Oper(&oand, &oator);
	GetAnswer(&oand, &oator, exp);
	printf("%s\n%d\n", exp, oand.and[oand.top-1]);
	system("pause");
	return 0;
}