HOW DOES PBP handle bits varables to bytes

[HenrikOlsson]

Hi,
I've looked at this and although I'm not that good at assembly here how I believe it works.
When you create a BIT variable the compiler allocates a BYTE of RAM and calls it PB1, your bit variable is then defined as the first bit in that byte. When you create a second BIT variable it gets defined as the second bit in the previously allocated BYTE (ie still only a single byte). When you create the 9th BIT variable a new BYTE is allocated (PB2) and so on. So, one BIT variable takes 1 BYTE but 8 bit variables still only takes ONE byte.

Basically, this is what it looks like in the assembly listing:

00081 PB01                            EQU     RAM_START + 019h
00083 PB02                            EQU     RAM_START + 01Ah
00092 #define _myBit1                  PB01, 000h
00093 #define _myBit2                  PB01, 001h
00094 #define _myBit3                  PB01, 002h
00095 #define _myBit4                  PB01, 003h
00096 #define _myBit5                  PB01, 004h
00097 #define _myBit6                  PB01, 005h
00098 #define _myBit7                  PB01, 006h
00099 #define _myBit8                  PB01, 007h
00100 #define _myBit9                  PB02, 000h

As you can see, for 9 BIT variables only TWO bytes of RAM is allocated - thankfully and as expected.

/Henrik.

[richard]

that's good to know , after misreading the manual at some stage I have been micro-managing bit vars needlessly for years .

[LinkMTech]

Good question and answers!
Like Richard, I have also been doing the same when it was already taken care of.