表达式求值（C实现，实现多括号，浮点数）---栈的实现以及运用。

//栈用单链表来实现

#include<stdio.h>
#include<stdlib.h>
#include<ctype.h>
#include<math.h>
#include<string.h>

struct Node1
{
char a;
struct Node1 *next;
};

struct Node2
{
double a;
struct Node2 *next;
};
typedef struct Node1 *Stack1;
typedef struct Node2 *Stack2;
Stack1 CreakStack1()
{
Stack1 S;
S = (Stack1)malloc(sizeof(Node1));
S->next = NULL;
return S;
}

Stack2 CreakStack2()
{
Stack2 S;
S = (Stack2)malloc(sizeof(Node2));
S->next = NULL;
return S;
}

void Push1(char c, Stack1 S)
{
Stack1 tmp;
tmp = (Stack1)malloc(sizeof(Node1));
tmp->a = c;
tmp->next = S->next;
S->next = tmp;

}

void Push2(double c, Stack2 S)
{
Stack2 tmp;
tmp = (Stack2)malloc(sizeof(Node1));
tmp->a = c;
tmp->next = S->next;
S->next = tmp;

}

void Pop1(Stack1 S)
{
Stack1 tmp ;
tmp = S->next;
S->next = tmp->next;
free(tmp);
}

void Pop2(Stack2 S)
{
Stack2 tmp;
tmp = S->next;
S->next = tmp->next;

}

char Top1(Stack1 S)
{
return(S->next->a);
}

double Top2(Stack2 S)
{
return(S->next->a);
}
bool Isempty1(Stack1 S)//这道题感觉没什么用！反正输入合法！
{
return S->next == NULL;
}
bool Isempty2(Stack2 S)
{
return S->next == NULL;
}

char Prior[8][8] =
{ // 运算符优先级表
// '+' '-' '*' '/' '(' ')' '#' '^'
/*'+'*/'>','>','<','<','<','>','>','<',
/*'-'*/'>','>','<','<','<','>','>','<',
/*'*'*/'>','>','>','>','<','>','>','<',
/*'/'*/'>','>','>','>','<','>','>','<',
/*'('*/'<','<','<','<','<','=',' ','<',
/*')'*/'>','>','>','>',' ','>','>','>',
/*'#'*/'<','<','<','<','<',' ','=','<',
/*'^'*/'>','>','>','>','<','>','>','>'
};

float Operate(float a, char c, float b)
{
switch (c)
{
case '+':return a + b;
case'-':return a - b;
case'*':return a*b;
case'/':return a / b;
case'^':return pow(a, b);
default:return 0;
}
}

char OP[] = { "+-*/()#^" };           //把可能遇到的8个操作数预先存到一个数组中

int In(char a)                       //判断字符是否为操作数
{
int flag = 0;
for (int i = 0; i < 8; i++)
if (a == OP[i]) { flag = 1; continue; }
return  flag;
}

int Ord(char a)                        //将操作数转化为坐标
{
for (int i = 0; i < 8; i++)
if (a == OP[i])return i;
}

char Precede(char a, char b)         //返回操作数的优先级
{
return(Prior[Ord(a)][Ord(b)]);
}

float Evaluate(char *Expression)
{
Stack1 S1 = CreakStack1();
Stack2 S2 = CreakStack2();
Push1('#', S1);
char cat[2] = "\0";
char digit[20] = "";  //从表达式数组中接受操作数部分
strcat(Expression, "#");
if (*Expression == '-');      //对第一个数是负数情况的处理；负数只会出现在字符串的第一个或者左括号的下一个哦！！！
Push2(0.0, S2);
while (!(*Expression == '#'&&Top1(S1) == '#'))
{
if (!In(*Expression))//操作数
{
cat[0] = *Expression;
strcat(digit, cat);
Expression++;

if (In(*Expression))
{
double Data = atof(digit);
Push2(Data, S2);
strcpy(digit, "\0");
}

}

else     //操作符
{
if (*Expression == '(')                    //遇到负数的情况的处理；
{
if (*++Expression == '-')
Push2(0.0, S2);                     //压入浮点数栈一个0；
Expression--;                           //回到左括号。
}
switch (Precede(Top1(S1), *Expression))
{
case'<':
Push1(*Expression, S1);
Expression++;
break;
case'=':
Pop1(S1);
Expression++;
break;
case'>':
char c = Top1(S1); Pop1(S1);
double a = Top2(S2); Pop2(S2);
double b = Top2(S2); Pop2(S2);
Push2(Operate(b, c, a), S2);
break;
}//switch
}//else
}//while
return(Top2(S2));

}

int main()
{
char s[200];
puts("请输入表达式（括号在英文状态下输入）");
scanf("%s", s);
puts("该表达式的值为");
printf("%f\n\n", Evaluate(s));
system("pause");

}