Question

我输入正则表达式并找到字符串最小长度的代码。例如：ab*(ab)+因此，要匹配此正则表达式的字符串的最小长度为3.但是在检查(a|b)时，我的代码会导致问题。它给出了分段错误。但它适用于(a|b)*。

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

 void main()
 {
struct node *regexp=null,*language=null;
char d,lokahead;
int i,j,k,countl=0,mincount=0,count=0;
printf("\nEnter the regular expression:");
for(i=0;;i++)
{
    scanf("%c",&d);
    if (d=='.')
        break;
    push(&regexp,d);
}
//traverse(&regexp);
mincount=check(&regexp);
printf("\nThe minimum size of the language should be %d",mincount);
printf("\nEnter the language:");
for(i=0;;i++)
{
    scanf("%c",&d);
    if (d=='.')
        break;
    countl++;
    push(&language,d);
}
}

---queue.h

#define null 0

 struct node{
        char data;
        struct node *next;
       };

 void push(struct node **head,char d);
 char pop(struct node **head);

int check(struct node **head)
{
    struct node *k;
    char d,d1;
    int count=0,mincount=0;
    if(*head!=null)
    {
        k=*head;
        while(k!=null)
        {
            d=k->data;
            if ((d>=97 && d<=122) || (d>=65 && d<=90))
            {
                k=k->next;
                d1=k->data;
                //printf("\n%c %d",d,mincount);
                if ((d1>=97 && d1<=122) || (d1>=65 && d1<=90) || (d1=='(') || (d1=='|'))
                    mincount++;
                /*if (d1=='+')
                {
                    mincount++;
                    k=k->next;
                }*/
            }
            else if (d=='*')
            {
                k=k->next;
            }
            else if (d=='+')
            {
                mincount++;             
                k=k->next;
            }
            else if (d=='(')    //checks for open (
            {
                k=k->next;  
                while(k->data!=')') //checks till close )
                {
                    if (k->data!='|')   //count the characters inside (...)                 
                        count++;
                    if ((k->data)=='|')         //counts the case (..a|b..) as 1 character
                        k=k->next;
                    k=k->next;      
                }
                printf("\nElements inside bracket %d",count);
                k=k->next;      //proceeds pointer so that we leave (...)
                if (k->data=='+')   //checks if we have (...)+ then the minimum charcters should be the number of char
                    mincount+=count;//inside the ()
                count=0;        
                k=k->next;      //increase the pointer to next element
            }
            else if (d=='|')
            {
                k=k->next;
                if (k->data!='+' || k->data!='*' || k->data!='(')
                    mincount++;
            }
        }
    }
    return mincount;
}

输出：

paku@ubuntu:./regex
Enter the regular expression:ab*(ab)+.

Elements inside bracket 2

The minimum size of the language should be 3

Enter the language:aab.

输出2：

paku@ubuntu:./regex
Enter the regular expression:ab*(a|b).

Segmentation fault.(core dumped)

Answer 1

由于它给出了分段错误，您的代码正在访问超出此程序范围的内存。检查指针

正则表达式使用（a | b）

1 个答案: