i**********e
发帖数: 1145 | 1 These are some original pointer problems from me. Try to solve without a
compiler's help. Post your answers here by replying to this post.
Assume you have the following linked list defined as:
struct Node {
Node(int x) : val(x), next(NULL) {}
int val;
Node *next;
};
And you have a printList function which prints out the content of a list.
What are the outputs for the following code segments? |
d*******d
发帖数: 2050 | |
r******n
发帖数: 170 | |
g**f
发帖数: 414 | 4 *p = new Node(4);
p = &((*p)->next); |
m*********2
发帖数: 701 | 5 that's what i thought too.
But i tried it with GCC.
it works.
The first replier (dinohound) got it all right.
【在 g**f 的大作中提到】
: *p = new Node(4);
: p = &((*p)->next);
|
i**********e
发帖数: 1145 | 6 dinohound，你的答案是完全正确的。
看来大家还是对 LEVEL 3 的感到有些混淆，那我先说我理解的，欢迎大家补充一下。
*p = new Node(4);
p = &((*p)->next);
Level 3 相信大家想问的就是应该编译会出现错误，因为 (*p)->next 不就是 NULL 吗
？然后相信很多人会对 &((*p)->next) 理解成 &(NULL)，而感到不解。
我给个简单些的例子，道理其实一样：
int *p = NULL;
那 &p 是什么呢？其实就是那个 p 变量的地址，p 变量储存的值到底是什么是没有关
系的。
同样的道理， p = &((*p)->next) 也就能理解成当前链表下一个节点的地址的地址。
所以 p 指向的东西（也就是下一个节点的地址），能通过 *p 来更改（使得下一个节
点的地址更改而指向一个新建的节点）。
一些常见面试题的答案与总结 -
http://www.ihas1337code.com |
g**f
发帖数: 414 | 7 Good explanation. Got it.
【在 i**********e 的大作中提到】
: dinohound，你的答案是完全正确的。
: 看来大家还是对 LEVEL 3 的感到有些混淆，那我先说我理解的，欢迎大家补充一下。
: *p = new Node(4);
: p = &((*p)->next);
: Level 3 相信大家想问的就是应该编译会出现错误，因为 (*p)->next 不就是 NULL 吗
: ？然后相信很多人会对 &((*p)->next) 理解成 &(NULL)，而感到不解。
: 我给个简单些的例子，道理其实一样：
: int *p = NULL;
: 那 &p 是什么呢？其实就是那个 p 变量的地址，p 变量储存的值到底是什么是没有关
: 系的。
|
d*******d
发帖数: 2050 | 8 thanks.
这种问题可以很磨人的.
【在 i**********e 的大作中提到】
: dinohound，你的答案是完全正确的。
: 看来大家还是对 LEVEL 3 的感到有些混淆，那我先说我理解的，欢迎大家补充一下。
: *p = new Node(4);
: p = &((*p)->next);
: Level 3 相信大家想问的就是应该编译会出现错误，因为 (*p)->next 不就是 NULL 吗
: ？然后相信很多人会对 &((*p)->next) 理解成 &(NULL)，而感到不解。
: 我给个简单些的例子，道理其实一样：
: int *p = NULL;
: 那 &p 是什么呢？其实就是那个 p 变量的地址，p 变量储存的值到底是什么是没有关
: 系的。
|
t******g
发帖数: 252 | 9 I don't understand how the second line in level one work. Can someone please
explain?
I think p is allocated as a pointer to pointer. But where it's pointed to is
not allocated. Why is that not a problem?
【在 i**********e 的大作中提到】
: These are some original pointer problems from me. Try to solve without a
: compiler's help. Post your answers here by replying to this post.
: Assume you have the following linked list defined as:
: struct Node {
: Node(int x) : val(x), next(NULL) {}
: int val;
: Node *next;
: };
: And you have a printList function which prints out the content of a list.
: What are the outputs for the following code segments?
|
m*********2
发帖数: 701 | 10 LZ不如谈下这low-level detail.
For example:
1) How is Node** p represented in the low level?
p, the pointer to pointer to a Node, is on the stack?
where is *p located then? on the stack or the heap?
My understanding is:
Node p ==> the struct is on the stack.
Node *p ==> the pointer is on the stack, the struct is on heap
Node **P ==> the pointer of pointer is on the stack... not sure where
the other two is going to be.
In addition, does the complier keep track of RTTI of p (in case of Node
**p)? Does the complier know that it's a pointer to pointer to Node? or
it is just a generic pointer?
2)
so, when Node.Next is assigned to NULL
Node.NEXT still has valid address, pointing to 0？
【在 i**********e 的大作中提到】
: dinohound，你的答案是完全正确的。
: 看来大家还是对 LEVEL 3 的感到有些混淆，那我先说我理解的，欢迎大家补充一下。
: *p = new Node(4);
: p = &((*p)->next);
: Level 3 相信大家想问的就是应该编译会出现错误，因为 (*p)->next 不就是 NULL 吗
: ？然后相信很多人会对 &((*p)->next) 理解成 &(NULL)，而感到不解。
: 我给个简单些的例子，道理其实一样：
: int *p = NULL;
: 那 &p 是什么呢？其实就是那个 p 变量的地址，p 变量储存的值到底是什么是没有关
: 系的。
|
|
|
t******g
发帖数: 252 | 11 Never mind. I get it.
p is a pointer to a pointer to Node. It was assigned pointer to head which
is itself a pointer to Node.
I think I was tired from a long day.
please
is
【在 t******g 的大作中提到】
: I don't understand how the second line in level one work. Can someone please
: explain?
: I think p is allocated as a pointer to pointer. But where it's pointed to is
: not allocated. Why is that not a problem?
|
i**********e
发帖数: 1145 | 12 Node *head = new Node(3);
Node **p = &head;
*p = new Node(4);
p = &((*p)->next);
*p = new Node(5);
Attached is an image to visualize the pointers for level three.
Remember, a pointer always occupy 4 bytes of memory no matter what type it
is. It stores nothing but the address of the data it is pointing to (which
is a 32 bit integer). The data type is only used to do pointer arithmetic (
ie, to calculate the number of bytes to jump over to the next element in an
array).
一些常见面试题的答案与总结 -
http://www.ihas1337code.com |
i**********e
发帖数: 1145 | 13 The pointer is always on the stack (it stores a 32-bit integer (the address
of the data it is pointing to)). The data it is pointing to can be on the
stack or the heap.
For example,
int *p = new int(3); // p is pointing to an int on the heap
int x = 3;
int *p = &x; // p is pointing to an int on the stack
The compiler knows that it's a pointer to pointer to Node. (you declare it
as Node **p). The pointer type is used for pointer arithmetic purpose. That'
s why a generic pointer (void *) could not perform pointer arithmetic.
Node.NEXT will have a valid address. Every variable must have a valid
address, right? Yes, and it is pointing to NULL. which means the pointer
itself has a value of 0. On the other hand, if the pointer is pointing to a
valid data, then the pointer itself would have the address of the valid data
(which must be NON-NULL(NON-ZERO)).
一些常见面试题的答案与总结 -
http://www.ihas1337code.com |
G**********s
发帖数: 70 | 14 ------------ LEVEL ZERO -----------------
Node *head = new Node(3);
Node *p = head;
p = new Node(4);
0.a) printList(head);
3
0.b) printList(p);
4
------------ LEVEL ONE ------------------
Node *head = new Node(3);
Node **p = &head;
*p = new Node(4);
1.a) printList(head);
4
1.b) printList(*p);
4
------------ LEVEL TWO ------------------
Node *head = new Node(3);
Node **p = &head;
*p = new Node(4);
*p = (*p)->next;
2.a) printList(head);
null
2.b) printList(*p);
null
------------ LEVEL THREE -----------------
Node *head = new Node(3);
Node **p = &head;
*p = new Node(4);
p = &((*p)->next);
*p = new Node(5);
3.a) printList(head);
4->5
3.b) printList(*p);
null |
G**********s
发帖数: 70 | 15 感谢你设计的题目帮我顺便又复习了一下，好玩！!
解pointer的题，要virtulize memory，脑子里就要生成一个像你画的那个图，这些题
就简单多了。 |
