Question

以下程序检查算术表达式，如a + b a-b，它给输出有效或无效;

%{
#include<stdio.h>
#include<stdlib.h>
int c,d,bo=0,bc=0;
%}
operand [a-zA-Z0-9]+
operator [+\-\/*]
%%
//the operand is one count higher than the operator if that fails then its is invalid eg a+b operand is two and operator is 1;

{operator} {d++;printf("%s is an operator \n",yytext);} 

{operand} {c=d+1;printf("%s is an operand \n",yytext);}

"(" {if(bc<=bo)bo++;}

")" {bc++;}

\n {if(bo==bc&&c>d){printf("valid exp");}else {printf("invalid exp");};exit(0);}
%%
void main(){
yylex();
}

我面临的问题是，当我检查一个++ b它说有效但是当我尝试+ b-和其他值如a）a + b（，（a + b（，+ a-b ++）给出我正确的输出。只有a ++ b和a - b或a + -b它才会失败。我有点卡住了。

这是\ n的if条件，我把它放在\ n bcz当我按下输入它给我输出并退出。

if(bo==bc && c>d)                   //c>d means if operand is greater than operator
    {  printf("valid exp");  }
      else {  
             printf("invalid exp"); }

Answer 1

我刚刚将c = d + 1改为c ++;这是一个逻辑错误产生错误，而不是检查操作数更大的操作符我添加1额外的操作数，它总是在++ b中评估为真

%{
#include<stdio.h>
#include<stdlib.h>
int c,d,bo=0,bc=0;
%}
operand [a-zA-Z0-9]+
operator [+\-\/*]
%%
//the operand is one count higher than the operator if that fails then its is invalid eg a+b operand is two and operator is 1;

{operator} {d++;printf("%s is an operator \n",yytext);} 

{operand} {c++;printf("%s is an operand \n",yytext);}

"(" {if(bc<=bo)bo++;}

")" {bc++;}

\n {if(bo==bc&&c>d){printf("valid exp");}else {printf("invalid exp");};exit(0);}
%%
void main(){
yylex();
}

Lex程序识别有效的算术表达式，并识别有效的标识和运算符

1 个答案: