我尝试为这个 bool 表达式求值问题编写最紧凑的 C 代码。这是我的最终代码:
编辑:已删除
这是添加的否定处理:
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; }
}
}
Testing:
#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 );
}