知識+ 直譯運算

//===================================================================
// 20090402 知識 +
// http://tw.knowledge.yahoo.com/question/question?qid=1609040211054
// 發問者 : http://tw.knowledge.yahoo.com/my/my?show=AF04167627
//===================================================================

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

double var[26] = {0};
char cd[256];
int cp;

char Op[]={ 0, '*', '/', '+', '-' };
int  On[]={ 0, 0x60, 0x61, 0x50, 0x51 };
int  lev=0;

int getN(double *n)
{
  *n = 0;
  int x=1;
  int e = -5;
  double d = 1.;
  if (cd[cp]=='-') { cp++; x=-1; };
  for( ;cd[cp]>='0' && cd[cp]<='9' || cd[cp]=='.'; cp++)
  {
    if (cd[cp]!='.')
    {
      e=0;
      if (d==1.)
      {
         *n = *n*10+(cd[cp]&0x0F);
         continue;
      }
      *n += d*(cd[cp]&0x0F);
    }
    else if (d<1) return -4;
    d /= 10.;
  }
  if (x<0) *n *= -1.;
  return e;
}

int get( double *n, int *op )
{
  int  i;
  while ( cd[cp]=='(' || cd[cp]==' ')
     if (cd[cp++]=='(') lev++;
  int p=cp;
  if ( (cd[cp]|0x20)>='a' && (cd[cp]|0x20)<='z')
  {  i=(cd[cp++]|0x20)-'a';  *n=var[i];  }
  else if ((i=getN(n))<0) return i;
  while (cd[cp]==')' || cd[cp]==' ')
     if (cd[cp++]==')')
        if (--lev<0) return -8;
  for(i=4; i>=0; i--)  if (cd[cp]==Op[i]) break;
  if (i<0) return -6;
  if (i==0 && lev>0) return -7;
  *op=(lev<<8) + On[i];
  cp++;
  return 0;
}

int oper( double *n1, int *o1 )
{
  double n2;
  int   o2;
  int   e=get( &n2, &o2);
  while( !e && (o2>>4) > ((*o1)>>4) ) e=oper( &n2, &o2 );
  if (e) return e;
  switch ( (*o1)&0xFF )
  {
    case 0:  *n1=n2; return 0;
    case 0x60: *n1*=n2; break;
    case 0x61: if (n2) *n1/=n2; else printf("/0錯誤!!\n"); break;
    case 0x50: *n1+=n2; break;
    case 0x51: *n1-=n2; break;
    default: return -3;
  }
  *o1=o2;
  return 0;
}

int parse()
{
  char ic[2][8] = { "quit", "print" };
  int c=3;
  int v=-1;
  int p=0;
  while(cd[p]==' ')p++;
  cp=p;
  while((cd[cp]|0x20)>='a'&& (cd[cp]|0x20)<='z')
     cd[cp++] |= 0x20;
  if (cp-p>1)  // 命令
  {
    cd[cp++] = '\0';
    for(c=1; c>=0 && strcmp( &cd[p], ic[c])!=0; c--);
    if (c!=1) return c;  // 0: 離開  -1: 命令錯誤
  }
  else if(cp-p==1)  // 變數
  {
    v = cd[p]-'a';
    while(cd[cp]==' ')cp++;
    if (cd[cp++]!='=') return -2;
    c=2;
  }
  else return (strlen(cd)==0)?3:-1;
  int op=0;
  double n=0.;
  lev = 0;
  p = oper(&n, &op);
  if (p<0) return p;
  if(c==1) printf("%f\n", n);
  if(v>=0) var[v] = n;
  return c; // 2:指定敘述 1:顯示敘述
}

int main()
{
  int c=1;
  printf("\n# Welcome!!\n");
  while(c!=0)
  {
    printf("]");
    gets(cd);
    c = parse();
    if (c<0) printf("*** (%d)Error!!\n", c);
  }
  printf("# Bye.\n");
  return 0;
}