c表达评估员

Question

好吧,我想在文本文件中有一个这样的字符串:

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

在将其解析为c程序并处理并正确设置变量之后,它将最终看起来像这样:

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

是否有任何有用的库用于评估表示为这样的一个字符串的表达式？我以为我可以用字符串作为参数调用一个perl程序,它可以很容易地返回结果,但不确定C中是否有一个库可以做到这一点,或者是否有任何已知的算法可以解决这样的表达？

编辑:我实际上正在寻找的东西会吐出这个表达的答案,也许解析是一个坏词.即1或0

在一个坚果壳中,它包含一堆随机表达式(已知是正确的格式),需要被评估为0或1.(上面计算结果为1,因为它导致(1和1).

Answer 1

您可以在程序中嵌入lua,然后调用它的解释器来计算表达式.

Answer 2

我试着为这个bool表达式评估问题编写最紧凑的C代码.这是我的最终代码:

编辑:删除

这是增加的否定处理:

编辑:添加测试代码

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 );
}

Answer 3

为这样的简单表达式滚动你自己的递归下降解析器是很容易的。

Answer 4

我有一个类似的程序来实现递归体面的解析器，所以我把它刷了一遍，就在这里。

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


int doOR(int pOprd1, int pOprd2) {
    if (pOprd1 == -1) return pOprd2;
    return pOprd1 || pOprd2;
}
int doAND(int pOprd1, int pOprd2) {
    if (pOprd1 == -1) return pOprd2;
    return pOprd1 && pOprd2;
}
int doProcess(char pOpert, int pOprd1, int pOprd2) {
    if (pOpert == '0') return pOprd2;
    if (pOpert == 'O') return doOR (pOprd1, pOprd2);
    if (pOpert == 'A') return doAND(pOprd1, pOprd2);
    puts("Unknown Operator!!!");
    exit(-1);
}
int* doParse(char pStr, int pStart) {
    char C;
    int  i        = pStart;
    int  Value    =  -1;
    char Operator = '0';
    for(; (C = pStr[i]) != 0; i++) {
        if (C == '0') { Value = doProcess(Operator, Value, 0); continue; }
        if (C == '1') { Value = doProcess(Operator, Value, 1); continue; }
        if (C == ' ') continue;
        if (C == ')') {
            int aReturn;
            aReturn = malloc(2*sizeof aReturn);
            aReturn[0] = Value;
            aReturn[1] = i + 1;
            return aReturn;
        }
        if (C == '(') {
            int * aResult = doParse(pStr, i + 1);
            Value = doProcess(Operator, Value, aResult[0]);
            i = aResult[1];
            if (pStr[i] == 0) break;
            continue;
        }
        if ((C == 'A') && ((pStr[i + 1] == 'N') && (pStr[i + 2] == 'D'))) {
            if ((Operator == '0') || (Operator == 'A')) {
                Operator = 'A';
                i += 2;
                continue;
            } else {
                puts("Mix Operators are not allowed (AND)!!!");
                exit(-1);
            }
        }
        if ((C == 'O') && (pStr[i + 1] == 'R')) {
            if ((Operator == '0') || (Operator == 'O')) {
                Operator = 'O';
                i += 1;
                continue;
            } else {
                puts("Mix Operators are not allowed (OR)!!!");
                exit(-1);
            }
        }
        printf("Unknown character: '%c (\"%s\"[%d])'!!!", C, pStr, i);
        exit(-1);
    }
    int* aReturn;
    aReturn = malloc(2*sizeof aReturn);
    aReturn[0] = Value;
    aReturn[1] = i;
    return aReturn;
}

这是一个测试代码：

int main(void) {
    char* aExpr   = "1";
    int*  aResult = doParse(aExpr, 0);
    printf("%s = %d\n", aExpr, ((int*)aResult)[0]);
    free(aResult);
    aExpr   = "0";
    aResult = doParse(aExpr, 0);
    printf("%s = %d\n", aExpr, ((int*)aResult)[0]);
    free(aResult);
    aExpr   = "1 AND 0";
    aResult = doParse(aExpr, 0);
    printf("%s = %d\n", aExpr, ((int*)aResult)[0]);
    free(aResult);
    aExpr   = "1 AND 1";
    aResult = doParse(aExpr, 0);
    printf("%s = %d\n", aExpr, ((int*)aResult)[0]);
    free(aResult);
    aExpr   = "0 OR 0 OR 0";
    aResult = doParse(aExpr, 0);
    printf("%s = %d\n", aExpr, ((int*)aResult)[0]);
    free(aResult);
    aExpr   = "1 OR 0 OR 0";
    aResult = doParse(aExpr, 0);
    printf("%s = %d\n", aExpr, ((int*)aResult)[0]);
    free(aResult);
    aExpr   = "1 OR 1 OR 0";
    aResult = doParse(aExpr, 0);
    printf("%s = %d\n", aExpr, ((int*)aResult)[0]);
    free(aResult);
    aExpr   = "(1 OR 0)";
    aResult = doParse(aExpr, 0);
    printf("%s = %d\n", aExpr, ((int*)aResult)[0]);
    free(aResult);
    aExpr   = "(0 OR 0)";
    aResult = doParse(aExpr, 0);
    printf("%s = %d\n", aExpr, ((int*)aResult)[0]);
    free(aResult);
    aExpr   = "((( 1 AND 0 AND 0) OR 1) AND ((0 OR 1) AND 1))";
    aResult = doParse(aExpr, 0);
    printf("%s = %d\n", aExpr, ((int*)aResult)[0]);
    free(aResult);
    puts("DONE!!!");
    return EXIT_SUCCESS;
}

这很有趣:-D。