我有一个名为“ startMyProc {num}”的进程。我希望该过程由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proc“ 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
}
每个线程都有自己的完整解释器，与程序中的所有其他解释器隔离（thread软件包的命令功能除外）。在所有线程中获取过程的最简单，最直接的方法是将其放在脚本文件中，然后将source其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 $num
}]
但是您会遇到另一个问题。变量也不能共享。那意味着您不会克服$num。您应该真正使脚本启动，然后thread::wait在最后完成。然后，您可以将thread::send其作为工作（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]
但是，如果您真的在考虑将任务发送给工作线程，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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