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Wait for user input with timeout (serial console) Digi Connect ME9210

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I've a Digi Connect ME9210 connected via serial console for debugging things. After boot I want to wait for 1-2s to give the user a chance to Enter some key (dont matter what key). After 1-2s without any key the programm must continues. How could I do this? scanf wait forever.
asked Mar 23 in NET+OS by ottelo New to the Community (3 points)

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2 Answers

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Best answer
it is already implemented in the serial dialog. When the board boots it displays a dialog and waits 5 seconds for a key press. Just look in dialog.c and do the same.
answered Mar 23 by LeonidM Veteran of the Digi Community (4,490 points)
selected Mar 29 by ottelo
0 votes
I copied the code from dialog.c -> checkKeyboardInput()

and put it into applicationStart(). But the function hangs until I enter a key.

The function content is:

static char waitForUserInputWithTimeout(unsigned long timeout)
unsigned int i;
char rc=0xff;

for (i=0; i < timeout; i++)
rc= (char) getchar();
if ((rc != 0xff) && (rc != '\n'))
return (rc);
return (rc);
}/* waitForUserInputWithTimeout */
answered Apr 5 by ottelo New to the Community (3 points)