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2 << -1 = 0?

BradDaBug · 2003-08-06T21:00:00

I''m trying to create a macro that''ll replace pow(2, n) using bitshifts. Here''s what it looks like so far: #define pow2(n) 2 << (n - 1) It works OK unless you pass 0, which then gets turned into 2 << -1. 2^0 is supposed to be 1, and 10 in binary is 2, and 10 shifted -1 to the left would be 01, which is 1, right? So why does 2 << -1 = 0 when it seems like it''d be 1? I''d hate to have to add a special case to the macro to return 1 if n is 0. Is there something simple I don''t understand?

General and Gameplay Programming Programming

Started by BradDaBug August 06, 2003 03:45 PM

10 comments, last by BradDaBug 20 years, 8 months ago

RenderTarget

398

August 06, 2003 08:21 PM

quote:Original post by Village Specialton
I don''t like #define. I prefer to use a variable. The books I read say to never use #define.

Scott Simontis
Big Joke: C#

Purists don''t want you to #define. But purists would rather write 20 lines of code where one would suffice, for the sake of purity. There are times when macros are appropriate.

I like pie.

[sub]My spoon is too big.[/sub]

DrPizza

160

August 06, 2003 09:00 PM

But this isn''t one of those times.

unsigned pow2(unsigned n) { return 1 << n; }
will be identical to the macro. You may need to add the keyword "inline" if your optimizer is picky.

char a[99999],*p=a;int main(int c,char**V){char*v=c>0?1[V]:(char*)V;if(c>=0)for(;*v&&93!=*v;){62==*v&&++p||60==*v&&--p||43==*v&&++*p||45==*v&&--*p||44==*v&&(*p=getchar())||46==*v&&putchar(*p)||91==*v&&(*p&&main(0,(char**)(--v+2))||(v=(char*)main(-1,(char**)++v)-1));++v;}else for(c=1;c;c+=(91==*v)-(93==*v),++v);return(int)v;}  /*** drpizza@battleaxe.net ***/

2 << -1 = 0?

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2 << -1 = 0?

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