后缀表达式计算

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#include <stdio.h>#include <stdlib.h>#include<string.h>#define MAXSIZE 30typedef struct{    char data[MAXSIZE];    int top;} SqStack;typedef struct{    double data[MAXSIZE];    int top;} SqStack1;//栈的初始化 void Stack_Init(SqStack *s){    s->top = 0;}void Stack_Init1(SqStack1 *s){    s->top = 0;}//判断栈是否为空int Stack_Empty(SqStack *s){    if (s->top>0)        return 1;    else        return 0;}//入栈void Stack_Push(SqStack *s, char ch){    if (s->top == MAXSIZE)    {        printf("栈满。无法入栈！！!\n");    }    else    {        s->data[s->top++] = ch;        //printf("  %c入栈! \n ", ch);    }}void Stack_Push1(SqStack1 *s, double n){    if (s->top == MAXSIZE)    {        printf("栈满。无法入栈！！!\n");    }    else    {        s->data[s->top++] = n;        //printf("  %f入栈! \n ", n);    }}void Stack_Pop(SqStack *s){    if (s->top == 0)        printf("栈空，无法出栈！！！！！！\n");    else    {        //printf("  %c出栈! \n ", s->data[s->top-1]);        s->top--;    }}void Stack_Pop1(SqStack1 *s){    if (s->top == 0)        printf("栈空，无法出栈！！！！！！\n");    else    {        //printf("  %f出栈! \n ", s->data[s->top - 1]);        s->top--;    }}//判断字符是否是操作符int Isoperationa(char ch){    int flag = 0;    if (ch == '+' || ch == '-' || ch == '*' || ch == '/')        flag = 1;    else        flag = 0;    return flag;}//判断符号优先级int priority(char ch){    int flag = 0;    switch (ch)    {    case'(':        flag = 0;        break;    case ')':        flag = 0;        break;    case '#':        flag = -1;        break;    case '+':        flag = 1;        break;    case '-':        flag = 1;        break;    case '*':        flag = 2;        break;    case'/':        flag = 2;        break;    default:        break;    }    return flag;}//验证表达式是否匹配void marry(char *infix){    int i = 0;    SqStack sta;    SqStack *s = &sta;    Stack_Init(s);    if (Isoperationa(infix[0]))    {        printf("非法输入！！！\n");        system("pause");        exit(-1);    }    while (infix[i] != '\0')    {        if (Isoperationa(infix[i]) && Isoperationa(infix[i + 1]))        {            printf("非法输入！！！\n");            system("pause");            exit(-1);        }        i++;    }    i = 0;    while (infix[i] != '\0')    {        if (infix[i] != '+' && infix[i] != '-' && infix[i] != '*' && infix[i] != '/'&&infix[i] != '('&&infix[i] != ')' && (infix[i] < '0' || infix[i] > '9')&&infix[i]!='.')        {            printf("非法输入！！！\n");            system("pause");            exit(-1);        }        i++;    }    i = 0;    while (infix[i] != '\0')    {        if (infix[i] == '(')            Stack_Push(s, infix[i]);        if (infix[i] == ')')        {            if (s->top == 0)            {                printf("括号不匹配！！！\n");                system("pause");                exit(-1);            }            else                Stack_Pop(s);        }        i++;    }    if (s->top != 0)    {        printf("括号不匹配！！！\n");        system("pause");        exit(-1);    }    if (Isoperationa(infix[i-1]))    {        printf("非法输入！！！\n");        system("pause");        exit(-1);    }}/*-------------------------------------  中缀表达式转后缀表达式  遍历中缀表达式中所有字符  1.若是数字直接输出成为后缀表达式的一部分；  2.若是符号则判断其与栈顶符号优先级是右括号  优先级不高于栈顶符号将栈顶元素依次出栈输出成为后缀表达式一部分  将当前符号入栈直到遍历完中缀表达式且栈空为止  3.输出后缀表达式  -------------------------------------*/void InfixToSuffix(char *infix, char *suffix){    SqStack sufstack;    SqStack *p = &sufstack;    Stack_Init(p);    int i = 0, j = 0;    while (infix[i]!='\0')    {        if (infix[i] >= '0'&&infix[i] <= '9'||infix[i]=='.')        {            suffix[j] = infix[i];            i++;            j++;        }        else if (Isoperationa(infix[i]))        {            suffix[j++] = ' ';            while (priority(infix[i]) <= priority(p->data[p->top - 1]))            {                suffix[j++] = p->data[p->top-1];                Stack_Pop(p);            }            Stack_Push(p, infix[i]);            i++;         }        else if (infix[i] == '(')        {            Stack_Push(p, infix[i]);            i++;        }        else if (infix[i] == ')')        {            while (p->data[p->top - 1] != '(')            {                suffix[j++] = p->data[p->top - 1];                Stack_Pop(p);            }            if (p->data[p->top - 1] = '(')            {                Stack_Pop(p);            }            i++;        }        else if (infix[i] == '.')            i++;    }    while (Stack_Empty(p))    {        suffix[j++] = p->data[p->top-1];        Stack_Pop(p);    }    suffix[j] = '\0';}//将后缀表达式中字符转换为数字double SuffixtoInt(char *suffix,int *i){    double num = 0.0;    int j = 0;    while (suffix[*i]<'0' || suffix[*i]>'9')        (*i)++;    while (suffix[*i] >= '0' && suffix[*i] <= '9')    {        num = num * 10 + suffix[*i] - '0';        (*i)++;    }    if (suffix[*i] == '.')    {        (*i)++;        while (suffix[*i] >= '0' && suffix[*i] <= '9')        {            j++;            num = num * 10 + suffix[*i] - '0';            (*i)++;        }    }    for (; j > 0; j--)    {        num = num / 10;    }    return num;}/*--------------------------------------------      后缀表达式计算结果      遍历后缀表达式遇到数字就进栈遇到符号就      将栈顶两个数字出栈进行运算并将计算结果      入栈直到获得结果  --------------------------------------------*/double Suffix_answer(char *suffix){    int i = 0;    double answer = 0.0;    SqStack1 Sta;    SqStack1 *p = &Sta;    Stack_Init1(p);    while (suffix[i] != '\0')    {        if (suffix[i] >= '0' && suffix[i] <= '9'||suffix[i]=='.')            Stack_Push1(p, SuffixtoInt(suffix, &i));        else if (Isoperationa(suffix[i]))        {            switch (suffix[i])            {                case '+':                     answer = p->data[p->top - 2] + p->data[p->top - 1];                     Stack_Pop1(p);                    Stack_Pop1(p);                    Stack_Push1(p, answer);                    i++;                    break;                case '-':                    answer = p->data[p->top - 2] - p->data[p->top - 1];                    Stack_Pop1(p);                    Stack_Pop1(p);                    Stack_Push1(p, answer);                    i++;                    break;                case '*':                    answer = p->data[p->top - 2] * p->data[p->top - 1];                    Stack_Pop1(p);                    Stack_Pop1(p);                    Stack_Push1(p, answer);                    i++;                    break;                case '/':                    answer = p->data[p->top - 2] / p->data[p->top - 1];                    Stack_Pop1(p);                    Stack_Pop1(p);                    Stack_Push1(p, answer);                    i++;                    break;            }        }        else            i++;    }      return answer;}int main(int argc, char *argv[]){    system("color 0A");    int i = 0;    int flag;    char ch;    char infix[30] = "";    char suffix[30]="";    printf("\t\t*****************************************************\n");    printf("\t\t*                           *\n");    printf("\t\t*                           *\n");    printf("\t\t*       计       算       器           *\n");    printf("\t\t*                           *\n");    printf("\t\t*                           *\n");    printf("\t\t*****************************************************\n\n");    while (1)    {        i = 0;        printf("\t\t请输入表达式：");        while ((ch = getchar()) != '\n')        {            infix[i++] = ch;        }        infix[i] = '\0';        marry(infix);        printf("\t\t中缀表达式：%s\n", infix);        InfixToSuffix(infix, suffix);        printf("\t\t后缀表达式：%s\n", suffix);        printf("\t\t计算结果：%f\n\n\n", Suffix_answer(suffix));        printf("\t\t是否继续 按n退出 按y继续！！\n\n\t\t");        fflush(stdin);        if ((flag = getchar())=='n')            exit(0);        if ((flag = getchar()) == 'y')        {            memset(infix, 0, sizeof(infix));            memset(suffix, 0, sizeof(suffix));        }    }    system("pause");    return 0;}
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