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[lua] Use the correct "controller" when "tool" param is specified in app.useTool()
Fix regression introduced in 1ec502d242ddddadf497c9738986edcaed55000b
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parent
6d2ddf877b
commit
7a74aef1a7
@ -293,15 +293,17 @@ int App_useTool(lua_State* L)
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lua_pop(L, 1);
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lua_pop(L, 1);
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// Select tool by name
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// Select tool by name
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const int buttonIdx = (params.button == tools::ToolLoop::Left ? 0: 1);
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auto activeToolMgr = App::instance()->activeToolManager();
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auto activeToolMgr = App::instance()->activeToolManager();
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params.tool = activeToolMgr->activeTool();
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params.tool = activeToolMgr->activeTool();
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params.ink = params.tool->getInk(params.button == tools::ToolLoop::Left ? 0: 1);
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params.ink = params.tool->getInk(buttonIdx);
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params.controller = params.tool->getController(params.button);
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params.controller = params.tool->getController(buttonIdx);
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type = lua_getfield(L, 1, "tool");
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type = lua_getfield(L, 1, "tool");
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if (type != LUA_TNIL) {
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if (type != LUA_TNIL) {
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if (auto toolArg = get_tool_from_arg(L, -1)) {
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if (auto toolArg = get_tool_from_arg(L, -1)) {
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params.tool = toolArg;
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params.tool = toolArg;
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params.ink = params.tool->getInk(0);
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params.ink = params.tool->getInk(buttonIdx);
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params.controller = params.tool->getController(buttonIdx);
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}
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}
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else
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else
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return luaL_error(L, "invalid tool specified in app.useTool() function");
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return luaL_error(L, "invalid tool specified in app.useTool() function");
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