Complete reverse Polish calculation

这是对之前的逆波兰式计算的修改，更改思路，使得可以处理多层函数和表达式嵌套的情况，如ln(ln(3))
这里是采用将数学函数和乘方号同样看成是一种优先级较高的操作符，进栈情况满足一般的逆波兰式，需要注意的就是小数点和乘方号的区分。
同样，这里采用数字栈和符号栈来同时表示一个表达式的逆波兰式，在数字栈中的符号位置使用MARK来标识，遇到MARK时，就去符号栈找对应的符号来计算即可。
中体思路还是和之前差不多的。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<cctype>
#define MAX 100
#define MARK 65535

typedef struct {
	char data[MAX];
	int top; 
}SqStack;

typedef struct {
	int prior[MAX];
	char opera[MAX];
	int top;
}Mark;

typedef struct{
	double data[MAX];
	char opera[MAX];
	int top;
	int otop;
}RPN;

SqStack *initStack(void);
void creatStack(SqStack *S);
void destroyStack(SqStack *S);
void printStack(SqStack *S);
void rpnStack(SqStack *S,RPN *rpn,Mark *mark);
void dealMark(RPN *rpn,Mark *mark,char ch,int pri);
void conduct(RPN *rpn);

void conduct(RPN *rpn)
{
	int i = 0;
	int j = 0;
	double temp[MAX];
	int p = 0;
	for(int i = 0;i<=rpn->top;i++)
	{
		if(rpn->data[i] == MARK)
		{
			switch (rpn->opera[p]){
				case '+' :{
					double a = temp[--j];
					double b = temp[--j];
					temp[j++] = a+b;
					break;
				}
				case '-':{
					double a = temp[--j];
					double b = temp[--j];
					temp[j++] = b-a;
					break;
				}
				case '*':{
					double a = temp[--j];
					double b = temp[--j];
					temp[j++] = a*b;
					break;
				}
				case '/':{
					double a = temp[--j];
					double b = temp[--j];
					temp[j++] = b/a;
					break;
				}
				case 'c':{
					double a = temp[--j];
					temp[j++] = cos(a);
					break;
				}
				case 's':{
					double a = temp[--j];
					temp[j++] = sin(a);
					break;
				}
				case 'o':{
					double a= temp[--j];
					temp[j++] = log10(a);
					break;
				}
				case 'n':{
					double a = temp[--j];
					temp[j++] = log(a);
					break;
				}
				case '^':{
					double a = temp[--j];
					double b = temp[--j];
					temp[j++] = pow(b,a);
					break;
				}
			}
			p++;
		}
		else
			temp[j++] = rpn->data[i];
	}
	printf("result = %.2f\n",temp[0]);
	//return temp[0];
}

void dealMark(RPN *rpn,Mark *mark,char ch,int pri)
{
	if(ch != '(' && ch != ')' && ch != ' ')
	{
		while(pri <= mark->prior[mark->top] && mark->opera[mark->top] != '(' && mark->top!=-1)
		{
			rpn->data[++rpn->top] = MARK;
			rpn->opera[++rpn->otop] = mark->opera[mark->top];
			mark->top--;
		}
		
		mark->opera[++mark->top] = ch;
		mark->prior[mark->top] = pri;
	}
	else if(ch == '(')
		mark->opera[++mark->top] = ch;
	else if(ch == ')')
	{
		while(mark->opera[mark->top]!='(')
		{
			rpn->data[++rpn->top] = MARK;
			rpn->opera[++rpn->otop] = mark->opera[mark->top];
			mark->top--;
		}
		mark->top--;
	}	
}

void rpnStack(SqStack *S,RPN *rpn,Mark *mark)
{	
	int t = 0;
	while(t<=S->top)
	{
		if(isdigit(S->data[t]))
		{
			char a[MAX];
			memset(a,0,sizeof(a));
			int ap = -1;
			
			int i = t;
			for(;isdigit(S->data[i]);i++);
			
			if(S->data[i] == '.')
			{
				i++;
				while(isdigit(S->data[i]))
				i++;
				for(int j = t;j<i;j++)
				a[++ap] = S->data[j];
				
				rpn->data[++rpn->top] = atof(a);
				t = i-1;
			}
			else
			{
				for(int j = t;j<i;j++)
				a[++ap] = S->data[j];
				
				rpn->data[++rpn->top] = atof(a);
				t = i-1;
			}
		}
		else if(S->data[t] == '+' || S->data[t] == '-')
			dealMark(rpn,mark,S->data[t],1);
		else if(S->data[t] == '*' || S->data[t] == '/')
			dealMark(rpn,mark,S->data[t],2);
		else if(S->data[t] == 'c' || S->data[t] == 's')
		{
			dealMark(rpn,mark,S->data[t],3);
			t = t+2;
		}
		else if(S->data[t] == 'l')
		{
			if(S->data[t+1] == 'o')
			{
				dealMark(rpn,mark,S->data[t+1],3);
				t = t+2;
			}
			else
			{
				dealMark(rpn,mark,S->data[t+1],3);
				t = t+1;
			}
		}
		else if(S->data[t] == '^')
			dealMark(rpn,mark,S->data[t],4);
		else if(S->data[t] == '(')
			dealMark(rpn,mark,S->data[t],5);
		else if(S->data[t] == ')')
			dealMark(rpn,mark,S->data[t],5);
		else
		continue ;
		
		t++;
	}
		
	while(mark->top>=0)
	{
		
		rpn->opera[++rpn->otop] = mark->opera[mark->top--];
		rpn->data[++rpn->top] = MARK;
	}
		
	conduct(rpn);
}

void creatStack(SqStack *S)
{
	char ch;
	while(scanf("%c",&ch)==1&&ch!='=')
	S->data[++S->top] = ch;	
}

void printStack(SqStack *S)
{
	int t = 0;
	while(t<=S->top)
	printf("%c",S->data[t++]);
	printf("\n");
}

void destroyStack(SqStack *S)
{
	free(S->data);
	free(S);
}

SqStack *initStack(void)
{
	SqStack *S = (SqStack *)malloc(sizeof(SqStack));
	if(!S)
	exit(0);
	
	S->top = -1;
	
	return S;
}


int main(void)
{
	freopen("逆波兰式.txt","r",stdin);
	
	SqStack *S = initStack();
	creatStack(S);
	printStack(S);
	
	Mark *mark = (Mark *)malloc(sizeof(Mark));
	for(int i = 0;i<MAX;i++)
	mark->prior[i] = -2;
	mark->top = -1;
	
	RPN *rpn = (RPN *)malloc(sizeof(RPN));
	rpn->otop = rpn->top = -1;
	
	rpnStack(S,rpn,mark);
	
	destroyStack(S);
	free(rpn->data);
	free(rpn->opera);
	free(rpn);
	free(mark->opera);
	free(mark->prior);
	free(mark);
	
	return 0;
}



//测试数据
//9.4+(3.27-1.05)*3.44+10/2.1+cos(0.52)+ln(ln(3))/log(5)+3^2=