本文介绍了在B树实现分段故障的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
任何人都可以请删除该段故障有所帮助。我工作的这个code一个星期仍无法调试此。这code是一个B树的实现。插入部分工作正常但在删除的分段错误。我无法调试它，任何人都可以请帮助？
Can anyone please help in removing this segmentation fault. I am working on this code for a week still unable to debug this. This code is a Btree implementation. The insertion part is working properly but there is an segmentation fault in deletion. I am unable to debug it, can anyone please help?
我给基于此链接输入（已经转换字母值ASCII值）
I have given the input based on this link (have converted alphabet value to ASCII value)http://cis.stvincent.edu/html/tutorials/swd/btree/btree.html
当我删除第一个 ^ h （相当于ASCII值）的正常工作，但是当我删除 T （相当于ASCII值），我会得到一个分段错误。
When I delete the first H (equivalent ASCII value) it works properly, but when I delete T (equivalent ASCII value) I will get a segmentation fault.
#include<stdio.h>
#include<stdlib.h>
#define M 5
struct node{
int n; /* n < M No. of keys in node will always less than order of B
tree */
int keys[M-1]; /*array of keys*/
struct node *p[M]; /* (n+1 pointers will be in use) */
}*root=NULL;
enum KeyStatus { Duplicate,SearchFailure,Success,InsertIt,LessKeys };
void insert(int key);
void display(struct node *root,int);
void DelNode(int x);
void search(int x);
enum KeyStatus ins(struct node *r, int x, int* y, struct node** u);
int searchPos(int x,int *key_arr, int n);
enum KeyStatus del(struct node *r, int x);
int input_array[20]= {65,67,71,78,72,69,75,81,77,70,87,76,84,90,68,80,82,88,89,83};
int main()
{
int choice, i,key = 11;
printf("Creation of B tree for node %d\n",M);
while(1)
{
printf("1.Insert\n");
printf("2.Delete\n");
printf("3.Search\n");
printf("4.Display\n");
printf("5.Quit\n");
printf("Enter your choice : ");
scanf("%d",&choice);
switch(choice)
{
case 1:
//printf("Enter the key : ");
//scanf("%d",&key);
//for(i=0;i<20;i++)
for(i=0;i<20;i++)
{
key = input_array[i];
insert(key);
}
//insert(key++);
//insert(key);
break;
case 2:
printf("Enter the key : ");
scanf("%d",&key);
DelNode(key);
break;
case 3:
printf("Enter the key : ");
scanf("%d",&key);
search(key);
break;
case 4:
printf("Btree is :\n");
display(root,0);
break;
case 5:
exit(1);
default:
printf("Wrong choice\n");
break;
}/*End of switch*/
}/*End of while*/
return 0;
}/*End of main()*/
void insert(int key)
{
struct node *newnode;
int upKey;
enum KeyStatus value;
value = ins(root, key, &upKey, &newnode);
if (value == Duplicate)
printf("Key already available\n");
if (value == InsertIt)
{
struct node *uproot = root;
root=malloc(sizeof(struct node));
root->n = 1;
root->keys[0] = upKey;
root->p[0] = uproot;
root->p[1] = newnode;
}/*End of if */
}/*End of insert()*/
enum KeyStatus ins(struct node *ptr, int key, int *upKey,struct node **newnode)
{
struct node *newPtr, *lastPtr;
int pos, i, n,splitPos;
int newKey, lastKey;
enum KeyStatus value;
if (ptr == NULL)
{
*newnode = NULL;
*upKey = key;
return InsertIt;
}
n = ptr->n;
pos = searchPos(key, ptr->keys, n);
if (pos < n && key == ptr->keys[pos])
return Duplicate;
value = ins(ptr->p[pos], key, &newKey, &newPtr);
if (value != InsertIt)
return value;
/*If keys in node is less than M-1 where M is order of B tree*/
if (n < M - 1)
{
pos = searchPos(newKey, ptr->keys, n);
/*Shifting the key and pointer right for inserting the new key*/
for (i=n; i>pos; i--)
{
ptr->keys[i] = ptr->keys[i-1];
ptr->p[i+1] = ptr->p[i];
}
/*Key is inserted at exact location*/
ptr->keys[pos] = newKey;
ptr->p[pos+1] = newPtr;
++ptr->n; /*incrementing the number of keys in node*/
return Success;
}/*End of if */
/*If keys in nodes are maximum and position of node to be inserted is
last*/
if (pos == M - 1)
{
lastKey = newKey;
lastPtr = newPtr;
}
else /*If keys in node are maximum and position of node to be inserted
is not last*/
{
lastKey = ptr->keys[M-2];
lastPtr = ptr->p[M-1];
for (i=M-2; i>pos; i--)
{
ptr->keys[i] = ptr->keys[i-1];
ptr->p[i+1] = ptr->p[i];
}
ptr->keys[pos] = newKey;
ptr->p[pos+1] = newPtr;
}
splitPos = (M - 1)/2;
(*upKey) = ptr->keys[splitPos];
(*newnode)=malloc(sizeof(struct node));/*Right node after split*/
ptr->n = splitPos; /*No. of keys for left splitted node*/
(*newnode)->n = M-1-splitPos;/*No. of keys for right splitted node*/
for (i=0; i < (*newnode)->n; i++)
{
(*newnode)->p[i] = ptr->p[i + splitPos + 1];
if(i < (*newnode)->n - 1)
(*newnode)->keys[i] = ptr->keys[i + splitPos + 1];
else
(*newnode)->keys[i] = lastKey;
}
(*newnode)->p[(*newnode)->n] = lastPtr;
return InsertIt;
}/*End of ins()*/
void display(struct node *ptr, int blanks)
{
if (ptr)
{
int i;
for(i=1;i<=blanks;i++)
printf(" ");
for (i=0; i < ptr->n; i++)
printf("%d ",ptr->keys[i]);
printf("\n");
for (i=0; i <= ptr->n; i++)
display(ptr->p[i], blanks+10);
}/*End of if*/
}/*End of display()*/
void search(int key)
{
int pos, i, n;
struct node *ptr = root;
printf("Search path:\n");
while (ptr)
{
n = ptr->n;
for (i=0; i < ptr->n; i++)
printf(" %d",ptr->keys[i]);
printf("\n");
pos = searchPos(key, ptr->keys, n);
if (pos < n && key == ptr->keys[pos])
{
printf("Key %d found in position %d of last dispalyednode\n",key,i);
return;
}
ptr = ptr->p[pos];
}
printf("Key %d is not available\n",key);
}/*End of search()*/
int searchPos(int key, int *key_arr, int n)
{
int pos=0;
while (pos < n && key > key_arr[pos])
pos++;
return pos;
}/*End of searchPos()*/
void DelNode(int key)
{
struct node *uproot;
enum KeyStatus value;
value = del(root,key);
switch (value)
{
case SearchFailure:
printf("Key %d is not available\n",key);
break;
case LessKeys:
uproot = root;
root = root->p[0];
free(uproot);
break;
}/*End of switch*/
}/*End of delnode()*/
enum KeyStatus del(struct node *ptr, int key)
{
int pos, i, pivot, n ,min;
int *key_arr;
enum KeyStatus value;
struct node **p,*lptr,*rptr;
if (ptr == NULL)
return SearchFailure;
/*Assigns values of node*/
n=ptr->n;
key_arr = ptr->keys;
p = ptr->p;
min = (M - 1)/2;/*Minimum number of keys*/
pos = searchPos(key, key_arr, n);
if (p[0] == NULL)
{
if (pos == n || key < key_arr[pos])
return SearchFailure;
/*Shift keys and pointers left*/
for (i=pos+1; i < n; i++)
{
key_arr[i-1] = key_arr[i];
p[i] = p[i+1];
}
return --ptr->n >= (ptr==root ? 1 : min) ? Success : LessKeys;
}/*End of if */
if (pos < n && key == key_arr[pos])
{
struct node *qp = p[pos], *qp1;
int nkey;
while(1)
{
nkey = qp->n;
qp1 = qp->p[nkey];
if (qp1 == NULL)
break;
qp = qp1;
}/*End of while*/
key_arr[pos] = qp->keys[nkey-1];
qp->keys[nkey - 1] = key;
}/*End of if */
value = del(p[pos], key);
if (value != LessKeys)
return value;
if (pos > 0 && p[pos-1]->n > min)
{
pivot = pos - 1; /*pivot for left and right node*/
lptr = p[pivot];
rptr = p[pos];
/*Assigns values for right node*/
rptr->p[rptr->n + 1] = rptr->p[rptr->n];
for (i=rptr->n; i>0; i--)
{
rptr->keys[i] = rptr->keys[i-1];
rptr->p[i] = rptr->p[i-1];
}
rptr->n++;
rptr->keys[0] = key_arr[pivot];
rptr->p[0] = lptr->p[lptr->n];
key_arr[pivot] = lptr->keys[--lptr->n];
return Success;
}/*End of if */
if (pos > min)
{
pivot = pos; /*pivot for left and right node*/
lptr = p[pivot];
rptr = p[pivot+1];
/*Assigns values for left node*/
lptr->keys[lptr->n] = key_arr[pivot];
lptr->p[lptr->n + 1] = rptr->p[0];
key_arr[pivot] = rptr->keys[0];
lptr->n++;
rptr->n--;
for (i=0; i < rptr->n; i++)
{
rptr->keys[i] = rptr->keys[i+1];
rptr->p[i] = rptr->p[i+1];
}/*End of for*/
rptr->p[rptr->n] = rptr->p[rptr->n + 1];
return Success;
}/*End of if */
if(pos == n)
pivot = pos-1;
else
pivot = pos;
lptr = p[pivot];
rptr = p[pivot+1];
/*merge right node with left node*/
lptr->keys[lptr->n] = key_arr[pivot];
lptr->p[lptr->n + 1] = rptr->p[0];
for (i=0; i < rptr->n; i++)
{
lptr->keys[lptr->n + 1 + i] = rptr->keys[i];
lptr->p[lptr->n + 2 + i] = rptr->p[i+1];
}
lptr->n = lptr->n + rptr->n +1;
free(rptr); /*Remove right node*/
for (i=pos+1; i < n; i++)
{
key_arr[i-1] = key_arr[i];
p[i] = p[i+1];
}
return --ptr->n >= (ptr == root ? 1 : min) ? Success : LessKeys;
}/*End of del()*/
这可能是什么问题？
What could be the problem?
推荐答案
不知道到底该如何工作，我可以说，你写的 P 矢量之外的
Without knowing exactly how this should work I can say that you write outside of the p vector on
for (i=0; i < rptr->n; i++)
{
lptr->keys[lptr->n + 1 + i] = rptr->keys[i];
// When you delete key 84, rptr->n is 4 at one point which takes you outside
// p[M]
lptr->p[lptr->n + 2 + i] = rptr->p[i+1];
}
Valgrind是一个很好用的工具，我用的valgrind -v --leak检查=全＆LT发现了这个问题，你的可执行＆GT;
这篇关于在B树实现分段故障的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！