关于多重指针问题

qq75437828 2010-02-02 11:34:06

typedef struct TREE_NODE{
TREE_TYPE value;
struct TREE_NODE *left;
struct TREE_NODE *right;
}TreeNode;

TreeNode **link;

link = ¤t–>left;

*link = current–>left;
请问一下，这两句意义上有什么不同？

原题如下：
void
delete( TREE_TYPE value )
{
TreeNode *current;
TreeNode **link;
/*
** First, locate the value. It must exist in the tree or this routine
** will abort the program.
*/
link = &tree;
while( (current = *link) != NULL && value != current–>value ){
if( value < current–>value )
link = ¤t–>left;
else
link = ¤t–>right;
}
assert( current != NULL );
/*
** We’ve found the value. See how many children it has.
*/
if( current–>left == NULL && current–>right == NULL ){
/*
** It is a leaf; no children to worry about!
*/
*link = NULL;
free( current );
}
else if( current–>left == NULL || current–>right == NULL ){
/*
** The node has only one child, so the parent will simply
** inherit it.
*/
if( current–>left != NULL )
*link = current–>left;
else
*link = current–>right;
free( current );
}
else {
/*
** The node has two children! Replace its value with the
** largest value from its left subtree, and then delete that
** node instead.
*/
TreeNode *this_child;
TreeNode *next_child;
this_child = current–>left;
next_child = this_child–>right;
while( next_child != NULL ){
this_child = next_child;
next_child = this_child–>right;
}
value = this_child–>value;
delete( value );
current–>value = value;
}
}

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某某9 2010-02-02

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link = ¤t–>left;
这个是返回link是TreeNode **类型的,link的值变了，变为current–>left的地址
. 注意->的优先级高于&
*link = current–>left
这个是返回*link是TreeNode *类型的。link的值没变，他指向的指针的值变了，和current–>left指向的地址一样

nobody@noone 2010-02-02

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link = ¤t–>left;

*link = current–>left;
请问一下，这两句意义上有什么不同？

link是二级指针设放的是一级指针P的地址
一级指针P放的是变量X的地址
变量X放数据

link = xxx 改变的是link的值，也就是将 link指向另一个P
*link = xxx 改变的是P的值也就是将P指向另一个X

回到
link = ¤t–>left;
link的值变了，变为current–>left的地址
*link = current–>left;
link的值没变，他指向的指针的值变了，和current–>left指向的地址一样

qq75437828 2010-02-02