發送列表指針到函數

Question

對於我正在做的關於C中指針的練習，我想通過將指針listStart發送到insertEntry函數以及我想要插入的結構來在鏈表的開始處插入結構。然而，對於這個當前的代碼，我無法做到這一點，因爲listStart指針並沒有攜帶它在主函數（列表的第一個結構）中指向的地址。

據我所知，我只是將指針本身複製到insertEntry函數中，因此它指向的地址被省略。這意味着我所得到的insertEntry函數中的listStart指針是一個空指針。

爲了解決這個問題，我嘗試將listStart作爲指針發送給insertEntry函數，但是，它只是給了我指向指向null的指針的指針。我試着將listStart的地址發送給那個不工作的函數，因爲它是向函數發送null的。

我有問題：是否有可能做到這一點，我只是錯過了什麼？或者這是不可能的？

謝謝先進。

// header to include standard input and output 
#include <stdio.h> 

struct entry 
{ 
    int value; 
    struct entry *next; 
}; 


// prototype for insertEntry function 
void insertEntry(struct entry *l, struct entry *i, struct entry *j); 

int main(void) 
{ 

    // declaration of array for linked list 
    struct entry list1 = { 1 }, list2 = { 2 }, list3 = { 3 }, list4 = { 4 }, list5 = { 5 }, insert = { 8 }; 
    struct entry *listStart = &list1; 

    // test to see if the value of the insert.value struct is correct 
    printf("insert.value = %i\n", insert.value); 


    // assign pointers in list.next to the next struct in the list to create a linked list 
    list1.next = &list2; 
    list2.next = &list3; 
    list3.next = &list4; 
    list4.next = &list5; 
    list5.next = (struct entry *) 0; 

    // print the linked list to make sure the pointers are going to the correct struct member 
    printf("Original list!\n"); 
    while (listStart != (struct entry *) 0) 
    { 
     printf ("%i\n", listStart->value); 
     listStart = listStart->next; 
    } 

    // send struct to change and struct to insert 
    insertEntry(listStart, &insert, &list1); 

    // restart the list from the beginning because in the last while loop the listStart was assigned to the null pointer. 
    listStart = &list1; 

    // print the new list to show what has been inserted and moved around 
    printf("New list!\n"); 
    while (listStart != (struct entry *) 0) 
    { 
     printf ("%i\n", listStart->value); 
     listStart = listStart->next; 
    } 

    return 0; 
} 

// function to insert a new struct in the list and redirect an old struct in the list 
void insertEntry(struct entry *l, struct entry *i, struct entry *j) 
{ 
    i->next = l; // this is assigning the mem add of the pointer is list2.next to that of insert.next 
    l = i; // this is assigning the mem add of i which would be insert.value to the pointer in list2.next 

} 

Answer 1

問題1

你正在改變其中listStart指向第一個while循環。在該循環結束時，listStart的值爲NULL。

我建議你使用一個不同的變量遍歷列表中的項目，並保持listStart指向列表的開始。

struct entry *iter = listStart; 
while (iter != NULL) 
{ 
    printf ("%i\n", iter->value); 
    iter = iter->next; 
} 

使用iter在這兩個循環，以保持listStart指向列表的開始。

問題2

調用insertEntry後，listStart不指向列表的開始。 list1也不是列表中的第一個條目。 insert是列表開始處的對象。通過

listStart = &insert; 

建議用於清理

線

l = i; 

替換行

listStart = &list1; 

在insertEntry是無用的。它在函數中沒有做任何事情。它也不會改變main中的任何內容。去掉它。

Answer 2

以下是你會怎麼做你的任務的例子：

void insertEntry(struct entry** pplistStart, struct entry *pNewEntry) 
{ 
    pNewEntry->next = *pplistStart; 
    *pplistStart = pNewEntry; 
} 

它可以被稱爲像

insertEntry(&listStart, insert); 

Answer 3

你有兩個問題： -

之一是： -

printf("Original list!\n"); 
while (listStart->next != (struct entry *) 0) 
{ 
    printf ("%i\n", listStart->value); 
    listStart = listStart->next; 
} 

// send struct to change and struct to insert 
insertEntry(listStart, &insert, &list1); 

在insertEntry

（）listStart現在都指向NULL，所以當你調用insertEntry（），你會得到段錯誤。

第二個問題是： -

你的邏輯在I-升通過地址>接下來是錯誤的。你應該在l-> next中分配i的地址。

void insertEntry(struct entry *l, struct entry *i, struct entry *j) 
{ 
    i->next = l; // this is assigning the mem add of the pointer is list2.next to that of insert.next 
    l = i; // this is assigning the mem add of i which would be insert.value to the pointer in list2.next 

} 

我修改了你的程序看看。

// header to include standard input and output 
#include <stdio.h> 

struct entry 
{ 
    int value; 
    struct entry *next; 
}; 


// prototype for insertEntry function 
void insertEntry(struct entry *l, struct entry *i, struct entry *j); 

int main(void) 
{ 

    // declaration of array for linked list 
    struct entry list1 = { 1 }, list2 = { 2 }, list3 = { 3 }, list4 = { 4 }, list5 = { 5 }, insert = { 8 }; 
    struct entry *listStart = &list1; 

    // test to see if the value of the insert.value struct is correct 
    printf("insert.value = %i\n", insert.value); 

    // assign pointers in list.next to the next struct in the list to create a linked list 
    list1.next = &list2; 
    list2.next = &list3; 
    list3.next = &list4; 
    list4.next = &list5; 
    list5.next = (struct entry *) 0; 

    // print the linked list to make sure the pointers are going to the correct struct member 
    printf("Original list!\n"); 
    while (listStart->next != (struct entry *) 0) 
    { 
     printf ("%i\n", listStart->value); 
     listStart = listStart->next; 
    } 

    // send struct to change and struct to insert 
    insertEntry(listStart, &insert, &list1); 

    // restart the list from the beginning because in the last while loop the listStart was assigned to the null pointer. 
    listStart = &list1; 

    // print the new list to show what has been inserted and moved around 
    printf("New list!\n"); 
    while (listStart != (struct entry *) 0) 
    { 
     printf ("%i\n", listStart->value); 
     listStart = listStart->next; 
    } 

    return 0; 
} 

// function to insert a new struct in the list and redirect an old struct in the list 
void insertEntry(struct entry *l, struct entry *i, struct entry *j) 
{ 
    l->next = i; // this is assigning the mem add of the pointer is list2.next to that of insert.next 
    l = i; // this is assigning the mem add of i which would be insert.value to the pointer in list2.next 

} 

Answer 4

後while (listStart != (struct entry *) 0)循環執行，listStart指向NULL和內部insertEntry，你到L istStart這是NULL分配刀片的next指針。

你可以嘗試以下兩種解決方案，

1. 內insertEntry變化

i->next = l

要

i->next = j

或

之前調用insertEntry點listStart指針list1

listStart = &list1; insertEntry(listStart, &insert, &list1);