[lwip-users] Re: [lwip] servicing timeouts

[Adam Dunkels]

Hi!

On Tuesday 13 November 2001 07.00, you wrote:
> In the lwip port to rtxc, i find that the timeouts is processed only when
> an application task issues a blocking call, such as sys_sem_wait. I begin
> to think now, If a design uses a single task, which has scheduled some
> timeouts and which has no potential blocking calls, then How and where will
> the timeouts be serviced ? Can someone explain this concept ?
>
> In the 6502 port, there is no OS and the stack runs as if it is a single
> thread of execution. In this case, as expected, the timeouts are serviced
> in the sys_main.

In a single-threaded system, timeouts must be serviced in a way similar to 
the 6502 port. I.e., by explicitly blocking the execution while waiting for a 
timeout to occur (this is how the sys_main() in the 6502 port works). When 
the timeout occurs, the corresponding timeout handler function is called.

/adam
-- 
Adam Dunkels <address@hidden>
http://www.sics.se/~adam
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