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“结构节点*”类型分配给“结构节点”类型时,类型不兼容 尽管已成功编译
struct Node *temp = list;
此行类似,但未显示错误。
答案 0 :(得分:5)
struct Node *temp = list;
与
相同struct Node *temp;
temp = list;
因此将错误行更改为
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一个常规变量。在表达式中使用时,它意味着您要取消引用指针。在声明期间无法取消引用它,这不会引起任何问题,因为无论如何这都是不确定的行为。
答案 1 :(得分:1)
首先出现错误的原因是...
*list = temp->next; //not *list it should be list
这样做可以编译您的代码,但是我认为您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
但是我认为您正在尝试进行线性搜索。 因此,您还需要将上述行更改为
temp = temp->next;
使代码按预期工作。
答案 2 :(得分:0)
因为参数list
的声明像
struct Node *list
然后在此语句中使用表达式*list
*list = temp->next;
具有类型struct Node
,而右侧操作数具有类型struct Node *
。
你必须写
list = temp->next;
但是无论如何要注意,当传递的列表为空时,该函数可以调用未定义的行为。并且应该在节点上搜索目标值,而不是仅检查根节点是否包含目标值。
最糟糕的是,该函数甚至不更改指向头节点的指针,因为该指针是通过值传递给该函数的。所以这句话
list = temp->next; // error here
将原始指针的副本更改为根节点,而不是在main中声明的原始指针本身。
该功能至少应定义为
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;
}
如果主要是您有声明
struct Node *list = NULL;
//...
然后必须像这样调用函数
remove_element( &list );
或类似的
if ( remove_element( &list ) )
{
puts( "A node was removed." );
}
如果函数仅做一件事,那就更好:删除节点。对于必须删除的值的提示应放在函数外部。在这种情况下,可以通过以下方式定义功能
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;
}
其输出为
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