trigger output bit by bit

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hello, i saw a demo code which change the function of trigger output from byte by byte to bit by bit, by using both of this symbol,
|= and &=
i wonder what is the function of it.. cause i cant find any solution from google~

Posted 26-Sep-12 20:53pm

arnoldxx

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**JF2015** · Accepted Answer · 2012-09-26T21:00:00

Solution 1

Hi,

to set a bit x, use this functionality:

number |= 1 << x;

to clear a bit x, use this code:

number &= ~(1 << x);

to toggle a bit x, use this code:

number ^= 1 << x;

See here for an explanation:
http://www.cprogramming.com/tutorial/bitwise_operators.html[^]

Posted 26-Sep-12 21:00pm

JF2015

Comments

arnoldxx 27-Sep-12 3:02am

thanks.. simple and nice explanation~ :)

arnoldxx 27-Sep-12 3:20am

sorry i m a bit confuse with this code

if(((CStatic*)pWnd)->GetBitmap() != m_bmp_Output_ON)
{
((CStatic*)pWnd)->SetBitmap(m_bmp_Output_ON);
wResult_4th_8Bits |= 0x80;
VTIO_OutpByte(wSelected_dwBaseAddr+0xcc, (unsigned char)wResult_4th_8Bits);
}
else
{
((CStatic*)pWnd)->SetBitmap(m_bmpOFF);
wResult_4th_8Bits &= 0x7f;
VTIO_OutpByte(wSelected_dwBaseAddr+0xcc, (unsigned char)wResult_4th_8Bits);
}

can u please explain it to me? thanks

JF2015 27-Sep-12 3:34am

The line wResult_4th_8Bits |= 0x80; basically only sets the 8th bit and the line wResult_4th_8Bits &= 0x7f; resets any bit above the 7th bit and leaves the lowest 7 bit unchanged, maybe just to convert a 32bit/16bit/8bit value to 7bit. Why this code performs this action, I can't tell you - this is not clear out of the context.

arnoldxx 27-Sep-12 3:38am

haha.. its okey, ^^ thanks for the explanation