RE: [Foxgui-users]general C question: explanation of timeout code in FXApp

[Daniel G. <bu...@li...>]

Mathew,

Well, it's perfectly legal C++... Memberwise comparison, see ARM
13.2.3.2

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.
.
.
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Surprised ? Just joking: see FXApp.cpp, line 1120:

static inline int operator<(const struct timeval& a,const struct
timeval& b){
  return (a.tv_sec<b.tv_sec) || (a.tv_sec==b.tv_sec &&
a.tv_usec<b.tv_usec);
  }

Regards,

Daniel

> -----Original Message-----
> From: fox...@li... 
> [mailto:fox...@li...] On Behalf 
> Of Mathew Robertson
> Sent: Wednesday, December 19, 2001 1:58 AM
> To: FOX Users
> Subject: [Foxgui-users]general C question: explanation of 
> timeout code in FXApp
> 
> 
> 
> Can someone explain to me how this works:
> 
> // Handle all past due timers
> gettimeofday(&now,NULL);
> while(timers){
>   register FXTimer* t=timers;
>   if(now < t->due) break;
>   timers=t->next;
>   if(t->target  && t->target->handle(this,MKUINT(t->message,
>                            SEL_TIMEOUT),&event)) refresh();
>   t->next=timerrecs;
>   timerrecs=t;
>   }
> 
> 
> Both 'now' and 't->due' are struct timeval's.  How can:
> 
> if (now < t->due) break;
> 
> work for a struct?
> 
> What am I missing here?
> 
> Mathew Robertson
> 
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