6-5 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 is supposed to return the value of the expression. If it is not a legal postfix expression, EvalPostfix must return a special value Infinity which is defined by the judge program.

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

代码:
ElementType EvalPostfix( char *expr ) {
    char t[10];
    double s[100],flag;
    int c = 0,k = 0;
    while(expr[k]) {
        int j = 0;
        while(expr[k] && expr[k] != ' ') t[j ++] = expr[k ++];
        t[j] = 0;
        if(t[0] >= '0' && t[0] <= '9' || t[1] && (t[0] == '-' || t[0] == '+')) {
            s[c] = 0;
            int i = 0;
            flag = 1;
            if(t[i] == '-') {
                flag = -1;
                i ++;
            }
            if(t[i] == '+') i ++;
            while(t[i] && t[i] != '.') {
                s[c] = s[c] * 10 + t[i ++] - '0';
            }
            if(t[i] == '.') {
                i ++;
                double d = 0.1;
                while(t[i]) {
                    s[c] += (t[i ++] - '0') * d;
                    d *= 0.1;
                }
            }
            s[c] *= flag;
            c ++;
        }
        else {
            if(c < 2) return Infinity;
            switch(t[0]) {
                case '+':s[c - 2] += s[c - 1];break;
                case '-':s[c - 2] -= s[c - 1];break;
                case '*':s[c - 2] *= s[c - 1];break;
                case '/': if(s[c - 1] == 0) return Infinity;
                          s[c - 2] /= s[c - 1];break;
            }
            c --;
        }
        while(expr[k] && expr[k] == ' ') k ++;
    }
    if(c > 1) return Infinity;
    return s[0];
}

 


免责声明!

本站转载的文章为个人学习借鉴使用,本站对版权不负任何法律责任。如果侵犯了您的隐私权益,请联系本站邮箱yoyou2525@163.com删除。



 
粤ICP备18138465号  © 2018-2025 CODEPRJ.COM