opration prior.cpp

来自「相当强悍」· C++ 代码 · 共 87 行

// Opration Prior.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <math.h>
#include <stdio.h>

char OPTR[10];
int f[6]={6,8,2,8,1};
int g[6]={4,7,9,2,1};
float OPND[10];
int OP=0,OR=1;
int m=0;

float Operat(char ch,float x,float y)//算符运算
 {
    switch (ch)
       {
	   case '+':  return x+y;
//	   case '-':  return x-y;
	   case '*':  return x*y;
//	   case '/':  return x/y;
        }
    printf("ERROR!\n");
	return 44;
}

int gg(char c)
{
  switch (c)
       {
           case '+':  return(0);
           case '*':  return(1);
           case '(':  return(2);
           case ')':  return(3);
           case '#':  return(4);
       }
    printf("ERROR!\n");
	return 44;
}

int Preced(int a,int b)
 {
    if(f[a]<g[b])
        return 1;
     else if(f[a]==g[b])
        return 2;
      else if(f[a]>g[b])
        return 3;
       else return 0;
 }

int main(int argc, char* argv[])
{
	float x,y;
	int t=0;
	char ch,c;
	OPTR[0]='#';
	printf("Please enter the expression and end with #:");//如  2*(3+4)#
	scanf("%c",&ch);
	while(OPTR[OR-1]!=OPTR[0]||ch!='#')
	 {
		if(ch<='9' && ch>='0')
			{ m=(ch-'0')+m*10;  scanf("%c",&ch);t=2;}//数字字符转换为数字,置t为2
		else if(!t&&ch!='(')  {printf("Too many opnd!\n");return 44;}//t为0或t不为( 
		else 
		{
			switch (Preced(gg(OPTR[OR-1]),gg(ch)))
			{	
				    case 1: if(t==2){OPND[OP++]=(float)m; m=0;}OPTR[OR++]=ch; scanf("%c",&ch);
					        if(ch=='(')t=1;
					        else t=0;  
							break;
					case 2: OR--; scanf("%c",&ch); break;
					case 3: if(t==2) {OPND[OP++]=(float)m;m=0;}
					        x=OPND[--OP];
					        y=OPND[--OP];c=OPTR[--OR];
					        OPND[OP++]=Operat(c,x,y);
					        if(ch!='#'&&ch!=')'){ OPTR[OR++]=ch;scanf("%c",&ch);t=0;}	   
							else t=1;  break;
			}
		}
	 }
	printf("%f",OPND[0]);
	return 0;
}