我編寫了此代碼以刪除單向鏈接列表中的第一個節點。將第一個節點從鏈接列表中刪除後錯誤結果
CreateLinkedList(node **headPtr)
{
int i;
node *pMyNode;
pMyNode = (node*)malloc(sizeof(node)); //create space for first node []
*headPtr=pMyNode;
for(i=0;i<10;i++)
{
pMyNode->element = i; //enter value [0]
printf("Value is %d addr is %p\n",pMyNode->element,pMyNode);
pMyNode->nextPtr = (node*)malloc(sizeof(node)); //[0]->[]->NULL
pMyNode = pMyNode->nextPtr;
}
pMyNode->nextPtr=NULL;
}
void PrintLinkedList(node **headPtr)
{
node *pMyNode;
int i;
pMyNode=*headPtr;
while(pMyNode)
{
printf("Value is %d addr is %p\n",pMyNode->element,pMyNode);
pMyNode = pMyNode->nextPtr;
}
}
void DeleteANode(node **headPtr)
{
node *pMyNode; //head->[]->[]->[]->NULL
pMyNode=*headPtr;
*headPtr=*headPtr->nextPtr;
free(pMyNode);
}
int main()
{
node *pNode;
CreateLinkedList(&pNode);
DeleteANode(&pNode);
PrintLinkedList(&pNode);
}
輸出我得到的是:
之前刪除
value is 0 addr is 8e75008
value is 1 addr is 8e75018
value is 2 addr is 8e75028
value is 3 addr is 8e75038
value is 4 addr is 8e75048
value is 5 addr is 8e75058
value is 6 addr is 8e75068
value is 7 addr is 8e75078
value is 8 addr is 8e75088
value is 9 addr is 8e75098
刪除
value is 0 addr is 8e75008 // This node should not be printed
value is 0 addr is 8e75018
value is 2 addr is 8e75028
value is 3 addr is 8e75038
value is 4 addr is 8e75048
value is 5 addr is 8e75058
value is 6 addr is 8e75068
value is 7 addr is 8e75078
value is 8 addr is 8e75088
value is 9 addr is 8e75098
請顯示更完整的代碼。您不會顯示節點正在創建/初始化的位置,或者添加到列表以及調用所有這些的代碼。 – OldProgrammer
DeletaANode意在刪除列表頭部的節點嗎?如果是這樣,爲什麼8e75018是不應該在您的示例中打印的節點; 8375008應該不是應該打印的節點? –
正確。我剛剛編輯了我的帖子。 – user968000