The ch_map[5] field in ble_gap_opt_ch_map_t to set channel map

[Mahendra Tailor]

Hi

SDK 15.2.0

ch_map[5] is 5 bytes long so 40 bits.

Can you please confirm if ....

ch_map[0] bit 0 corresponds to channel 0

ch_map[0] bit 7 corresponds to channel 7

ch_map[1] bit 0 corresponds to channel 8

ch_map[1] bit 7 corresponds to channel 15

ch_map[2] bit 0 corresponds to channel 16

ch_map[2] bit 7 corresponds to channel 23

etc

Look forward to hearing from you.

Kind regards
Mahendra

[Mahendra Tailor]

Hi
Thank you for the confirmation.
Actually the link you provide is the output from the documentation in the code which I had checked.

Unless I am not seeing it, there actually is no information in the link you provide that maps the bits in the 5 byte array to the actual channel numbers. And it is the logical channel numbers that it seems to link to, not the physical channel numbers (37.38.39 are adverts but spread across the spectrum in physical space) so again that needs to be clarified.

May I suggest that this information is updated in the documentation that is in the code.

In the meantime I will convert this query into a public one

Kind regards
Mahendra

[AndreasF]

Hi Mahendra.

I think you are suggesting documenting the endian-ness nature of ch_map in that structure. If that is correct, then I do not think that we need to. As a little-endian device, that is implicit that the byte order you mentioned will be that way and not in any other way.

The description of the bit you mentioned in your actual query is breakdown of that member in the little-endian way. Which is implicit.

if you even feel that the below member is not clear to you as to which bit refers to which channel, then I can send a message to the team to review it again.

  uint8_t ch_map[5];                      /**< Channel Map (37-bit). */

- Andreas

[Mahendra Tailor]

Thank you.