C防爆Ex pression计算器 [英] c expression Evaluator

问题描述

好吧可以说我有一个字符串，像这样一个文本文件：

Okay lets say I have a string such as this in a text file:

((( var1 AND var2 AND var3) OR var4) AND ((var5 OR var6) AND var7))

解析到这一点的C程序和增值经销商进行处理并正确设置它最终会看起来像这样经过：

after parsing this into the c program and the vars are handled and set correctly it will end up looking something like this:

((( 1 AND 0 AND 0) OR 1) AND ((0 OR 1) AND 1))

是否有任何有用的库外面，用于评估前pressions被重新psented因为像这样的字符串$ P $？我想我可以只调用一个Perl程序的字符串作为参数，将能够轻松地返回结果，但不知道是否有在C库，这样做，或是否有解决任何已知的算法如前pressions？

Are there any useful libraries out there for evaluating expressions that are represented as one string like this? I was thinking I could just call a perl program with the string as an argument that would be able to return the result easily but wasn't sure if there was a library in C that did this, or if there are any known algorithms for solving such expressions?

编辑：什么我其实在寻找的东西，会吐出一个答案前pression，也许是地分析一个坏词。即1或0

What i'm actually looking for is something that would spit out an answer to this expression, maybe parse was a bad word. i.e. 1 or 0

在一个螺母壳其包含一串需要进行评估，以0或1（随机前pressions（已经知道是在正确的格式）的一个文件以上的计算结果为1，因为它导致（1和1）。

In a nut shell its a file containing a bunch of random expressions (already known to be in the right format) that need to be evaluated to either 0 or 1. (above evaluates to 1 because it results in (1 AND 1).

推荐答案

我试着写最紧凑的C code代表这个布尔前pression评价问题。这是我最后的code：

I tried to write the most compact C code for this bool expression evaluation problem. Here is my final code:

编辑：删除

下面是添加否定处理：

编辑：测试$ C $下加入

test code added

char *eval( char *expr, int *res ){
  enum { LEFT, OP1, MID, OP2, RIGHT } state = LEFT;
  enum { AND, OR } op;
  int mid=0, tmp=0, NEG=0;

  for( ; ; expr++, state++, NEG=0 ){
    for( ;; expr++ )
         if( *expr == '!'     ) NEG = !NEG;
    else if( *expr != ' '     ) break;

         if( *expr == '0'     ){ tmp  =  NEG; }
    else if( *expr == '1'     ){ tmp  = !NEG; }
    else if( *expr == 'A'     ){ op   = AND; expr+=2; }
    else if( *expr == '&'     ){ op   = AND; expr+=1; }
    else if( *expr == 'O'     ){ op   = OR;  expr+=1; }
    else if( *expr == '|'     ){ op   = OR;  expr+=1; }
    else if( *expr == '('     ){ expr = eval( expr+1, &tmp ); if(NEG) tmp=!tmp; }
    else if( *expr == '\0' ||
             *expr == ')'     ){ if(state == OP2) *res |= mid; return expr; }

         if( state == LEFT               ){ *res  = tmp;               }
    else if( state == MID   && op == OR  ){  mid  = tmp;               }
    else if( state == MID   && op == AND ){ *res &= tmp; state = LEFT; }
    else if( state == OP2   && op == OR  ){ *res |= mid; state = OP1;  }
    else if( state == RIGHT              ){  mid &= tmp; state = MID;  }
  }
}

测试：

#include <stdio.h> 

void test( char *expr, int exprval ){
  int result;
  eval( expr, &result );
  printf("expr: '%s' result: %i  %s\n",expr,result,result==exprval?"OK":"FAILED");
}
#define TEST(x)   test( #x, x ) 

#define AND       && 
#define OR        || 

int main(void){
  TEST( ((( 1 AND 0 AND 0) OR 1) AND ((0 OR 1) AND 1)) );
  TEST( !(0 OR (1 AND 0)) OR !1 AND 0 );
}

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