表达式求值（C实现，实现多括号，浮点数）---栈的实现以及运用。

刚学完栈的时候写的，主要锻炼下栈的C实现吧!

//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");}