what's the size of one node? and where is the data saved?

Question

As the question implies,what's the size of usual node:

struct node{
   int data;
   node* next;
};

If the struct holds the data and the address of next node, what's the size of the pointer address(not what it points to)? Is there a difference if the next points to NULL?

where is the data(which is not a pointer) saved, stack or heap, if I made "new node":

node* linkednode= new node;

Are the answers change if I made class node?

That's implementation defined, we can't give you an answer, only your compiler on your platform can. — nikolas, Oct 02 '13 at 08:04
Think about the size and alignment of the members. Those put bounds on the size and alignment of the whole class. (Consider `node[N]`.) — Kerrek SB, Oct 02 '13 at 08:06

score 1 · Answer 1 · answered Oct 02 '13 at 08:07

1

You can easily check those using the sizeof() function applying it not only to your int and pointer variables but for the node itself. Usually ints and pointers are 4 bytes on 32 bit platforms, so this is implementation defined and might vary on other platforms. The total size of node might be equal to 8 but might not due to the cause specified above and\or data alignment.

answered Oct 02 '13 at 08:07

Alexandru Barbarosie

2,952
3
24
46

"Usually" on 32-bit platforms. – johnsyweb Oct 02 '13 at 08:08

Some programmer dude · Answer 2 · 2013-10-02T08:10:53.580

0

The size of a structure is fixed, no matter what the member values are. In this case it's probably either 8 or 16 bytes (depending on if you are on a 32 or 64 bit platform). This will be reported if you check e.g. sizeof(node).

The difference in size depending on 32 or 64 bit platforms, is because the pointer. Pointers are 32 bits (i.e. 4 bytes) on 32 bit platforms, and 64 bits (i.e. 8 bytes) on 64 bit platforms. The int type is usually 32 bits (4 bytes) on all platforms, but may be different on embedded systems or mainframes.

The reason it's not 12 bytes (4 + 8) on 64 bit platforms is because of alignment. The compiler will make sure that the pointer is on a good alignment to speed up access to that member.

edited Oct 02 '13 at 08:10

answered Oct 02 '13 at 08:04

Some programmer dude

400,186
35
402
621

I don't think it can sensibly be 12 bytes, unless there are some really weird alignment relaxations. – Kerrek SB Oct 02 '13 at 08:07
@KerrekSB You're probably right. Updated answer, with a small explanation of why. – Some programmer dude Oct 02 '13 at 08:09
@JoachimPileborg: I think the explanation is slightly off the point. Even if you reversed the member order, the size would be 16, not 12. – Kerrek SB Oct 02 '13 at 08:10
A lot of older PC's had 16 bit `int`; on them, size would have been 4 or 6, depending on the size of a pointer. This might still be the case on some embedded systems. – James Kanze Oct 02 '13 at 08:13
2

@KerrekSB Probably qualifies as "really weird", but on a Unisys mainframe, the size would be 12: a 6 byte int and a 6 byte pointer. – James Kanze Oct 02 '13 at 08:14
@JamesKanze: Yes on both counts, and +1 for the example :-) – Kerrek SB Oct 02 '13 at 08:16

what's the size of one node? and where is the data saved?

2 Answers2