Noobie Question re MCP23S17 Peter Oakes C# driver

I guess I should be able to answer that

the address << 1 will shift the address one space to the left leaving bit 0 for use as the r/w bit as per the specifications fo rthe chip (Page 8 of the PDF here: - http://ww1.microchip.com/downloads/en/DeviceDoc/21952b.pdf .

The address register (3 bits)  gives the chip the ability to have 8 devices  on one spi chip select. the R/W bit is for Read or Write operations, this is the same if the chip was the I2C variety

logical OR works very simply

each BIT is logicaly or's with the other parameter so for example if i have the following (The example is using arbitary values, not from the code sample)

0 1 0 0 1 0 1 0 <-- base address

1 1 0 0 1 0 0 0 <-- address left shifted one bit

1 1 0 0 1 0 1 0 <-- answer

so in summary, 1|0 = 1, 0|1 = 1, 1|1 = 1, 0|0 = 0

a very simple mechanism of taking a number and merging into it a bit mask for some set of a sub register to result int he complete value to write to the register. could this be simplifed yes but in this case it is part of a generic routine and if you simplified it for a single application and configuration it would no longer be generic

I guess I should be able to answer that

the address << 1 will shift the address one space to the left leaving bit 0 for use as the r/w bit as per the specifications fo rthe chip (Page 8 of the PDF here: - http://ww1.microchip.com/downloads/en/DeviceDoc/21952b.pdf .

The address register (3 bits)  gives the chip the ability to have 8 devices  on one spi chip select. the R/W bit is for Read or Write operations, this is the same if the chip was the I2C variety

logical OR works very simply

each BIT is logicaly or's with the other parameter so for example if i have the following (The example is using arbitary values, not from the code sample)

0 1 0 0 1 0 1 0 <-- base address

1 1 0 0 1 0 0 0 <-- address left shifted one bit

1 1 0 0 1 0 1 0 <-- answer

so in summary, 1|0 = 1, 0|1 = 1, 1|1 = 1, 0|0 = 0

a very simple mechanism of taking a number and merging into it a bit mask for some set of a sub register to result int he complete value to write to the register. could this be simplifed yes but in this case it is part of a generic routine and if you simplified it for a single application and configuration it would no longer be generic

Thanks to both of you for putting me straight. I think the newbie mistake was reading 0x40 as decimal not a hexadecimal.

Just to confirm. Structure of address byte is - BaseAddr followed by three bit device address followed by R/W bit (which actually comes from BaseAddr in that BaseAddR has a trailing 1 (for Read) which is passed through the mask after the left shift opens the mask for that bit.

i.e

BaseAddW                  0100 0000

BaseAddW                  0100 0001

DeviceAddr               0000 0111 (for device 7, is use this since device 0 is trivial and not very informative visually)

Shifted DeviceAddr   0000 1110

OR'd result  Write             0100 1110

OR'd result Read              0100 1111

I wanted to mark you both as correct as it was a combination of the answers which gave me the clues I needed. I had to give it to Peter in the end as it was his original code. Sorry Mike but your help was equally appreciated.

No problem. You could always mark mine as helpful

Mike

Anytime you see 0x in front of a number, it will indicate hex.

Mike

How's that!

Super!!!

Mike