Post by thread:to invert a single bit

>Does anyone know of a low instruction count method of inverting a >single bit in a register. For example inverting bit 0 in the PortB >register. I'm making this up as I go, but what about: movlw 0x01 xorwf portb Does that work? Any bits except for bit 0 are preserved (if it's a 0 xored w/ 0, you get 0; if it's a 1 xored w/ 0, you get 1), and bit 0 is toggled ( a 1 xored w/ 1 is 0, a 0 xored w/ 1 is 1). Right? Dave Johnson

Hi David, >Hello, > Does anyone know of a low instruction count method of inverting a >single bit in a register. For example inverting bit 0 in the PortB >register. A word of caution, changing a bit in an I/O port is inherently different than changing any other bit. All methods to change a bit (even using the bsf or bcf instructions) will first read the register, modify it, and then write it back out. When an I/O port is read you don't read what is programmed on an output line, but what is being read on an input line. As the output transistor supplies about 100 ohms of resistance it is possible that external loading or noise can cause a different value to be read than is programmed. This can cause bits other than the one you are toggling to be changed, or the one you are attempting to toggle to not change. To avoid this problem I map the output register values into an internal RAM location. By modifying the internal RAM value and then writing it out this problem is avoided. Example: PortBMap equ 20 ; define a map for the RAM. ; code fragment to invert bit 0 of PORTB, using PortBMap. movlw 1 ; toggle the lowest bit. xorwf PortBMap,f ; in the map. movf PortBMap,w ; now copy the map movwf PORTB ; to PORTB. Chip

On Wed, 18 Nov 1998, Dave Johnson wrote: > >Does anyone know of a low instruction count method of inverting a > >single bit in a register. For example inverting bit 0 in the PortB > >register. > I'm making this up as I go, but what about: > > movlw 0x01 > xorwf portb > > Does that work? Any bits except for bit 0 are preserved (if it's a 0 > xored w/ 0, you get 0; if it's a 1 xored w/ 0, you get 1), and bit 0 is > toggled ( a 1 xored w/ 1 is 0, a 0 xored w/ 1 is 1). Right? > > Dave Johnson > > It's o.k. However, one can get problem if try to invert a port bit directly. Why? Because of all port modifying operations (except outputting) are read-modify-write. If you have for instance a relay and you pull it down, you emit a bit 0 for it. Then if you xor another bit in that port, the level on the relay pin will be read. However it will bit 1 because of a to-be-pulled down relay is tied to Vdd. So if you emit the read bit unchanged, it will change the previous state. The solution? I use shadow registers, and maintain their contents. Imre