如何在C中交换两个结构

Question

I was looking at a code , written in C that swaps places of wto numbers and then two structs. I couldn't understand the second one :

#define SWAP(a,b) do {NODE *t = (a) ; (a) = (b) ; (b) = t;}

why does it work?? when I declare the t pointer of some struct "Node" so I directly point on a , then all data from b is being transfered to a , and b points to a as well... so I get that they both point on the same object(struct) .

If I write : Node t = *a instead , shouldn't it make it work? or i'm mistaken..

thanks!!

Answer 1

In this particular case, you are not transferring any data between the structures. You are just working on pointers which hold the addresses of the structures.

First, you create a pointer t which starts to point to whatever a was pointing to. Then, you modify pointer a so that it points to whatever b was pointing to. And finally, you modify b to point to whatever t is pointing to or IOW to what a was pointing before.

So, to sum up: you aren't moving any data between the structures *a and *b (where * means dereferencing the pointer) but just swapping the pointers a and b .

Answer 2

This code implies that both a and b are pointers of type NODE * . Suppose the two structs are the following: realA is a struct of type NODE and is pointed to by a; realB is also of type NODE and is pointed to by b. Now when you write this:

NODE *t = a;

here the temporary variable t, being a pointer and assigned to a, points to realA. Now when you see

a = b;

a no longer points to realA, but gets the address stored in b, so now a points to realB. And finally and similarily,

b = t;

infers that now b gets the pointer stored in t, which is, in turn, the address of realA.

All in all, only the pointers' contents have been changed; the actual data was untouched.

Answer 3

显而易见的猜测是，它们不会交换指针指向的东西，而只是交换指针本身。

Answer 4

This code swaps two pointers and doesn't copy any other data. It seems like a bad macro name; SWAP_NODE_PTR would be better.