Sar*_*ran 2 multithreading tcl
我有一个名为“startMyProc {num}”的过程。我希望这个 proc 被两个不同的线程调用并等待两个线程完成。我尝试了给出的有效解决方案。我想访问 startMyProc 中的全局变量并调用另一个过程“startMyAnotherProc {num}”。如何才能做到这一点?
package require Thread
global myVar
set myVar false
set id1 [thread::create -joinable {
source sample.tcl
thread::wait
}]
set id2 [thread::create -joinable {
source sample.tcl
thread::wait
}]
set num 1
thread::send -async $id1 [list startMyProc $num]
set num 2
thread::send -async $id2 [list startMyProc $num]
thread::join $id1
thread::join $id2
My sample.tcl looks like this,
proc startMyProc { num } {
global myVar
puts $myVar
puts "Opening $num"
after 2000
puts "Opening $num"
after 2000
puts "Opening $num"
after 2000
startMyAnotherProc $myVar
return
}
proc startMyAnotherProc { num } {
puts "Opening Another Proc: $num"
after 2000
puts "Opening Another Proc: $num"
after 2000
return
}
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每个线程都有自己完整的解释器,与程序中的所有其他解释器隔离(thread
包的命令功能除外)。在所有线程中获取过程的最简单、最直接的方法是将它放在一个脚本文件中,然后source
作为线程启动脚本的一部分:
set t1 [thread::create -joinable {
source myProcedures.tcl
startMyProc $num
}]
set t2 [thread::create -joinable {
source myProcedures.tcl
startMyProc $num
}]
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不过你会遇到另一个问题。变量也不能共享。这意味着你不会克服$num
。你真的应该让脚本启动然后thread::wait
在最后做。然后,您可以thread::send
完成工作(并在构建脚本时获得正确的替换)。
set t1 [thread::create -joinable {
source myProcedures.tcl
thread::wait
}]
set t2 [thread::create -joinable {
source myProcedures.tcl
thread::wait
}]
thread::send -async $t1 [list startMyProc $num]
thread::send -async $t2 [list startMyProc $num]
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但是,如果您真的考虑将任务发送到工作线程,则应该查看线程池 ( tpool
) 支持;这是很多以扩大更容易。
# Make the thread pool
set pool [tpool::create -initcmd {
source myProcedures.tcl
}]
# Sent the work into the pool as distinct jobs
set job1 [tpool::post $pool [list startMyProc $num]]
set job2 [tpool::post $pool [list startMyProc $num]]
# Wait for all the jobs in the pool to finish
set waitingfor [list $job1 $job2]
while {[llength $waitingfor] > 0} {
tpool::wait $pool $waitingfor waitingfor
}
# Get results now with tpool::get
# Dispose of the pool
tpool::release $pool
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