6-1 Evaluate Postfix Expression （25 分）

Write a program to evaluate a postfix expression. You only have to handle four kinds of operators: +, -, x, and /.

Format of functions:

ElementType EvalPostfix( char *expr );

where expr points to a string that stores the postfix expression. It is guaranteed that there is exactly one space between any two operators or operands. The function EvalPostfix

Sample program of judge:

#include <stdio.h>
#include <stdlib.h>

typedef double ElementType;
#define Infinity 1e8
#define Max_Expr 30   /* max size of expression */

ElementType EvalPostfix( char *expr );

int main()
{
    ElementType v;
    char expr[Max_Expr];
    gets(expr);
    v = EvalPostfix( expr );
    if ( v < Infinity )
        printf("%f\n", v);
    else
        printf("ERROR\n");
    return 0;
}

/* Your function will be put here */

 

Sample Input 1:

11 -2 5.5 * + 23 7 / -

Sample Output 1:

-3.285714

Sample Input 2:

11 -2 5.5 * + 23 0 / -

Sample Output 2:

ERROR

Sample Input 3:

11 -2 5.5 * + 23 7 / - *

Sample Output 3:

ERROR

Special thanks to Sirou Zhu (朱思柔) for providing a valuable test case.

ElementType EvalPostfix( char *expr ){
    double a[100]; //儲存資料，模擬棧
    char b[100]; //從表示式取字元
    int i=0,k=0; //記步
    while (expr[i]!='\0') {
        int j=0;
        while (expr[i]==' ') { //去掉表示式前面的空格和轉換成數字後的空格
            i++;
        }
        while((expr[i]=='.')||(expr[i]<='9'&&expr[i]>='0')||(expr[i]=='-'&&(expr[i+1]!=' '&&expr[i+1]!='\0'))){
            b[j]=expr[i];
            j++;
            i++;
        }//讀取字元
        if(expr[i]==' '||(k==0&&expr[i+1]=='\0')) {
            a[k]=atof(b);
            memset(b,0,sizeof(b));
            i++;
            k++;
        }//將字元轉化為double類資料
        else if(expr[i]!=' '&&(expr[i]>'9'||expr[i]<'0')&&(expr[i+1]==' '||expr[i+1]=='\0')){
            switch (expr[i]) {
                case '+':
                    a[k-2]=a[k-1]+a[k-2];
                    break;
                case '-':
                    a[k-2]=a[k-2]-a[k-1];
                    break;
                case '*':
                    a[k-2]=a[k-1]*a[k-2];
                    break;
                case '/':
                    if (a[k-1]==0) {
                        return Infinity;
                    }
                    a[k-2]=a[k-2]/a[k-1];
                    break;
                default:
                    break;
            }
            i++;
            k--;
        }//檢查到運算子，運算
    }
    if (k==0) {
        return Infinity;
    }
    return a[k-1];
}