Message ID | 1414064030-11029-3-git-send-email-lersek@redhat.com |
---|---|
State | New |
Headers | show
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diff --git a/IntelFrameworkModulePkg/Universal/BdsDxe/FrontPage.c b/IntelFrameworkModulePkg/Universal/BdsDxe/FrontPage.c index 219f691..c093416 100644 --- a/IntelFrameworkModulePkg/Universal/BdsDxe/FrontPage.c +++ b/IntelFrameworkModulePkg/Universal/BdsDxe/FrontPage.c @@ -931,7 +931,15 @@ ShowProgress ( EFI_GRAPHICS_OUTPUT_BLT_PIXEL Color; if (TimeoutDefault == 0) { - return EFI_TIMEOUT; + // + // this amounts to a poll -- 1 * 100ns timeout + // + Status = WaitForSingleEvent (gST->ConIn->WaitForKey, 1); + + if (Status == EFI_TIMEOUT) { + return EFI_TIMEOUT; + } + return HandleKeyPress (); } DEBUG ((EFI_D_INFO, "\n\nStart showing progress bar... Press any key to stop it! ...Zzz....\n"));
When PlatformBdsEnterFrontPage() is called with a zero TimeoutDefault value, PlatformBdsEnterFrontPage() --> ShowProgress() returns EFI_TIMEOUT immediately, without looking at the keyboard at all. This is slightly incorrect, because a keypress might already be pending at that time. Change ShowProgress() such that it polls the keyboard once even if TimeoutDefault equals zero, using a 100 nanosecond event timeout. (The UEFI specification explicitly allows a zero TriggerTime argument in gBS->SetTimer() -- meaning "next timer tick" for TimerRelative --, but passing zero to gBS->SetTimer() would require a reorganization of the WaitForSingleEvent() utility function. So let's just use a 100ns timeout here: it is indistinguishable from "next timer tick" for the purposes of polling the keyboard for an already pending keypress.) Contributed-under: TianoCore Contribution Agreement 1.0 Signed-off-by: Laszlo Ersek <lersek@redhat.com> --- Notes: v2: - new in v2 IntelFrameworkModulePkg/Universal/BdsDxe/FrontPage.c | 10 +++++++++- 1 file changed, 9 insertions(+), 1 deletion(-)