Question

我写了这个函数：

void something(struct node* head1, struct node* head2)
{
    statement 1 .......
    statement 2 .......
    so on.........
    // Want to make list2 (head1) be the same as list1 (head1):
    head2=head1
}

但这不会改变head2。它在功能中有效，但一旦回到主程序就没有，为什么？

Answer 1

似乎你想要head1和head2都指向同一个链表，你的代码的问题是你传递参数作为值调用，这就是为什么它没有反映你需要传递参数的函数通过Call By Pointer（参考）。试试这种方式

struct Node
{
int info;
Node *next;   
}

main{
Node * head1,*head2;

// call like this

something(&head1,&head2);

}

something(Node **temphead1, Node **temphead2){

//here you can use 

//(*temphead1)

//and 

//(*temphead2)

//to perform operation 

//for example you can display list item

while((*temphead1)!=null){

printf("%d",(*temphead1)->info);

(*temphead1)=(*temphead1)->next;
}

while((*temphead2)!=null){

printf("%d",(*temphead2)->info);

(*temphead2)=(*temphead2)->next;

}
// below line will do what you were looking for

(*temphead2)=(*temphead1);

// now you can check head1 =head2 

}

功能，链表。将一个链表复制到另一个链表

1 个答案: