You could avoid the entire temporary variable with the XOR method:
Code:#include <stdio.h> #define swap(a,b) a^=b^=a^=b int main(void) { int i, j; i = 5; j = 10; printf("i: %d, j: %d\n", i, j); swap(i, j); printf("i: %d, j: %d", i, j); return 0; }
You could avoid the entire temporary variable with the XOR method:
Code:#include <stdio.h> #define swap(a,b) a^=b^=a^=b int main(void) { int i, j; i = 5; j = 10; printf("i: %d, j: %d\n", i, j); swap(i, j); printf("i: %d, j: %d", i, j); return 0; }
Yeah, that too (or why it's a bad idea) is also in the clc FAQ
http://www.eskimo.com/~scs/C-faq/faq.html
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
why does everyone uses thestuff ???Code:do{...}while(0);
two simple brackets do just fine {...}
That fails when a and b are equal.Originally Posted by SirNot
My way, unfortunatly C99 or C++
good stuff about this is that it works for any type, including unions, structs or even C++ classes.Code:#define swap(a,b) {\ char t[sizeof(a)];\ memcpy(t,&a,sizeof(a));\ memcpy(&a,&b,sizeof(a));\ memcpy(&b,t,sizeof(a));\ }
Last edited by xErath; 06-12-2005 at 06:29 PM.
I believe it's stated in the standard somewhere that you should use that. And actually, it's:
That way when someone using the macro uses it, they can tack on a semicolon at the end and be correct.Code:do{...}while(0)
>Nice, but can you spot the bug in Dave's solution?
>why does everyone uses the
>two simple brackets do just fine {...}
Or, follow the link.Code:#include <stdio.h> #define why() { puts("because"); } int main(void) { int i = 42; if ( i ) why(); else puts("oh"); return 0; }
>That fails when a and b are equal.
It UB, pure and simple.
Last edited by Dave_Sinkula; 06-12-2005 at 07:49 PM.
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
Going back to the OP's original question, (which I'm still not too clear about) GCC provides an operator called typeof() - would this help? I appreciate the non-portability implications.
>>GCC provides an operator called typeof()
It does? I could't find anything on it.
Even if that operator exists, I wouldn't suggest using it. If you can't figure out a solution to your problem that works in most compilers, then you need to put your thinking cap back on.
>Even if that operator exists, I wouldn't suggest using it. If you can't figure out a solution to your problem that works in most compilers, then you need to put your thinking cap back on.
I do agree. It looks like a nice feature, but I would never use it personally. A quick search on google shows that MS Visual C++ supports it too. Maybe one day it'll become part of the standard, then we can all use it.
