从指针到指针的不兼容类型错误

Question

void remove_element(struct Node *list)
{   
    struct Node *temp = list;
    printf("Enter the element value you want to remove");
    int value;
    scanf("%d",&value);
    if(temp->data == value){ //first node is to be deleted
        *list = temp->next; // error here
        free(temp);
    }
}

错误：从类型 'struct Node *' 分配给类型 'struct Node' 时类型不兼容| 虽然已经编译成功

 struct Node *temp = list;

这一行很相似，但没有显示错误。

Answer 1

struct Node *temp = list;

与

相同

struct Node *temp;
temp = list;

所以把错误的那一行改成

list = temp->next;

请注意，* 的含义因上下文而略有不同。在声明中，它表示您要声明一个指针而不是常规变量。当在表达式中使用时，它意味着您想要取消引用指针。在声明期间无法取消引用它，这不会导致任何问题，因为无论如何这都是未定义的行为。

Answer 2

作为参数list声明为

struct Node *list

然后表达式*list用于此语句

*list = temp->next;

的类型为 struct Node 而右侧操作数的类型为 struct Node *.

你必须写

list = temp->next;

但无论如何要注意，当传递的列表为空时，函数可以调用未定义的行为。并且它应该在一个节点中搜索目标值，而不是只检查头节点是否包含目标值。

最糟糕的是函数甚至不更改指向头节点的指针，因为指针是按值传递给函数的。所以这个声明

    list = temp->next; // error here

将原始指针的副本更改为头节点，而不是在 main 中声明的原始指针本身。

函数的定义至少要像

int remove_element( struct Node **list )
{   
    printf( "Enter the element value you want to remove: " );
    int value;
    
    int success = scanf( "%d", &value ) == 1;

    if ( success )
    {
        while ( *list != NULL && ( *list )->data != value )
        {
            list = &( *list )->next;
        }

        success = *list != NULL;

        if ( success )
        {
            struct Node *tmp = *list;
            *list = ( *list )->next;
            free( tmp );
        }
    }

    return success;
}

如果在 main 中你有一个声明

struct Node *list = NULL;
//...

那么函数必须像

那样调用

remove_element( &list );

或类似

if ( remove_element( &list ) )
{
    puts( "A node was removed." );
}

另外，如果该函数只做一件事会更好：删除一个节点。对于必须删除的值的提示应该放在函数之外。在这种情况下，可以按以下方式定义函数

int remove_element( struct Node **list, int value )
{   
    while ( *list != NULL && ( *list )->data != value )
    {
        list = &( *list )->next;
    }

    int success = *list != NULL;

    if ( success )
    {
        struct Node *tmp = *list;
        *list = ( *list )->next;
        free( tmp );
    }

    return success;
}

这是一个演示程序。

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

struct Node
{
    int data;
    struct Node *next;
};
    
    
int push_front( struct Node **list, int data )
{
    struct Node *temp = malloc( sizeof( struct Node ) );
    int success = temp != NULL;

    if ( success )
    {       
        temp->data = data;
        temp->next  = *list;
        *list = temp;
        
    }

    return success;
}

void clear( struct Node **list )
{
    while ( *list != NULL )
    {
        struct Node *temp = *list;
        *list = ( *list )->next;
        free( temp );
    }
}

void display( const struct Node *list )
{
    for ( const struct Node *current = list; current != NULL; current = current->next ) 
    {
        printf( "%d -> ", current->data );
    }
    
    puts( "null" );
}

int remove_element( struct Node **list, int value )
{   
    while ( *list != NULL && ( *list )->data != value )
    {
        list = &( *list )->next;
    }

    int success = *list != NULL;

    if ( success )
    {
        struct Node *tmp = *list;
        *list = ( *list )->next;
        free( tmp );
    }

    return success;
}

int main(void) 
{
    struct Node *list = NULL;
    
    const int N = 10;
    
    for ( int i = N; i != 0; i-- )
    {
        push_front( &list, i );
    }
    
    display( list );
    
    int value = 1;
    
    if ( remove_element( &list, value ) )
    {
        printf( "The element with the value %d was removed.\n", value );
        
    }

    display( list );

    value = 10;
    
    if ( remove_element( &list, value ) )
    {
        printf( "The element with the value %d was removed.\n", value );
        
    }

    display( list );

    value = 5;
    
    if ( remove_element( &list, value ) )
    {
        printf( "The element with the value %d was removed.\n", value );
        
    }

    display( list );

    clear( &list );
    
    display( list );

    return 0;
}

它的输出是

1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 10 -> null
The element with the value 1 was removed.
2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 10 -> null
The element with the value 10 was removed.
2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null
The element with the value 5 was removed.
2 -> 3 -> 4 -> 6 -> 7 -> 8 -> 9 -> null
null

Answer 3

首先，您出现错误的原因是...

*list = temp->next; //not *list it should be list

这样做你的代码会编译，但我认为你会得到意想不到的结果。因为：

list=temp->next // This will make the pointer to constantly point to the head

但我认为您正在尝试进行线性搜索。因此，您还需要将上面的行更改为

temp = temp->next;

让您的代码按预期工作。

从指针到指针的不兼容类型错误

Incompatible types error from pointer to pointer

c

struct

linked-list

singly-linked-list

function-definition