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fix: hang by start_maintenance #1734

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fix: hang by start_maintenance #1734

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14 changes: 11 additions & 3 deletions RimeWithWeasel/RimeWithWeasel.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -116,10 +116,18 @@ void RimeWithWeaselHandler::Initialize() {

LOG(INFO) << "Initializing la rime.";
rime_api->initialize(NULL);
if (rime_api->start_maintenance(/*full_check = */ False)) {
m_disabled = true;
HANDLE hMutex =
CreateMutexW(NULL, FALSE, L"Global\\WeaselStartMaintenanceMutex");
if (hMutex) {
if (WaitForSingleObject(hMutex, 0) == WAIT_OBJECT_0) {
if (rime_api->start_maintenance(/*full_check = */ False)) {
rime_api->join_maintenance_thread();
m_disabled = true;
}
ReleaseMutex(hMutex);
}
CloseHandle(hMutex);
}

RimeConfig config = {NULL};
if (rime_api->config_open("weasel", &config)) {
if (m_ui) {
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18 changes: 17 additions & 1 deletion WeaselServer/WeaselServer.cpp
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Expand Up @@ -61,7 +61,23 @@ int WINAPI _tWinMain(HINSTANCE hInstance,

hRes = _Module.Init(NULL, hInstance);
ATLASSERT(SUCCEEDED(hRes));

HANDLE hMutex =
OpenMutexW(SYNCHRONIZE, FALSE, L"Global\\WeaselStartMaintenanceMutex");
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@lotem lotem Nov 16, 2025

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發生了什麼事？
算法服務進程不是只能運行一個實例嗎。

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@fxliang fxliang Nov 16, 2025

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这还没有到原来的mutex位置，比如/q参数的执行就不会运行到那个单实例的位置

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@lotem lotem Nov 17, 2025

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/q 让当前实例以 client 身份向正在运行的算法服务进程发信，那就不需要因为 start_maintenance 互斥吧。加了锁，会不会反而不能发信让主进程退出了？

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@fxliang fxliang Nov 17, 2025

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/q 让当前实例以 client 身份向正在运行的算法服务进程发信，那就不需要因为 start_maintenance 互斥吧。加了锁，会不会反而不能发信让主进程退出了？

目前试的状态是如果服务正在做start_maintenance，会堵塞住不响应其他业务。因此这个pr是要在start_maintenance 的时候标记mutex，其他尝试这种时候ipc通信的情况直接短路跳过，避免客户端堵塞卡死。

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@lotem lotem Nov 17, 2025

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好像是，有這個問題。
因爲在主線程等待嗎？

我認爲比較理想的修改是像用戶主動發起的部署過程一樣，算法服務能夠回應 IPC，返回「維護中」的狀態。

start_maintenance 有回調，weaselserver 應該有機會設置「維護中」的 Mutex。這條路竟然沒走通嗎。
看來是因爲主線程在等待部署結束，還不能處理 IPC 的原因。再研究研究。

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@lotem lotem Nov 17, 2025

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大概就是 join_maintenance_thread 造成的。
阻塞式初始化部署的目的是讓第一次按鍵就能出字；因爲通常軟件安裝時 SharedData 已經有部署好的詞典，這次初始部署很快，用戶能等。

小狼毫的情況又有所不同，安裝過程中已經在 UserData 做了部署，理論上已經能保證第一次按鍵時，輸入法的數據可用。初始部署是不是可以不等待，直接進入 IPC 循環？如果真的配置變化，就處於「維護中」狀態。

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@fxliang fxliang Nov 17, 2025

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潜在可能有一种情况，用户未知方式修改了方案，然后未知原因的服务退出，这个时候前端尝试拉起服务会卡住（部署堵塞）。试过不join thread，那就算加了mutex都避不了卡。在回调里处理mutex和init里处理效果一样

if (hMutex) {
DWORD wr = WaitForSingleObject(hMutex, 0);
if (wr == WAIT_TIMEOUT) {
CloseHandle(hMutex);
return 0;
}
if (wr == WAIT_OBJECT_0 || wr == WAIT_ABANDONED) {
ReleaseMutex(hMutex);
}
CloseHandle(hMutex);
}
if (hMutex) {
ReleaseMutex(hMutex);
CloseHandle(hMutex);
}
if (!wcscmp(L"/userdir", lpstrCmdLine)) {
CreateDirectory(WeaselUserDataPath().c_str(), NULL);
WeaselServerApp::explore(WeaselUserDataPath());
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14 changes: 14 additions & 0 deletions WeaselTSF/WeaselTSF.cpp
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Expand Up @@ -235,6 +235,20 @@ void WeaselTSF::_Reconnect() {
static unsigned int retry = 0;

bool WeaselTSF::_EnsureServerConnected() {
// if maintenancing, return false
HANDLE hMutex =
OpenMutexW(SYNCHRONIZE, FALSE, L"Global\\WeaselStartMaintenanceMutex");
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@lotem lotem Nov 16, 2025

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建立連接也需要檢查嗎？

算法服務中持續較長時間的維護工作處於單獨的線程，主線程仍可處理與前端的通信。
應該可以建立連接，並讓前端獲知算法服務處於維護模式。

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@fxliang fxliang Nov 16, 2025

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试过似乎是不行的，所以才每次都耗些微秒来检查这个

if (hMutex) {
DWORD wr = WaitForSingleObject(hMutex, 0);
if (wr == WAIT_TIMEOUT) {
CloseHandle(hMutex);
return false;
}
if (wr == WAIT_OBJECT_0 || wr == WAIT_ABANDONED) {
ReleaseMutex(hMutex);
}
CloseHandle(hMutex);
}
if (!m_client.Echo()) {
_Reconnect();
retry++;
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