ESP32 Forum

Tried to use various ESP32 diagnostics functions. Most of them work as expected, but:
ESP.getChipModel() return pointer to an unterminated string. Or the rest of the string is model-specific? In my case, it is "ESP32-D0WDQ6"
The same is about esp_get_idf_version(). It returns: v5.5.1-710-g8410210c9a
Function ESP.getSdkVersion() returns mess: 1061169513
Looks like I cannot detect which WiFI version is supported...
How to properly use these functions?

In my case, it is "ESP32-D0WDQ6"

That's the full name/variant of the SOC.

The same is about esp_get_idf_version(). It returns: v5.5.1-710-g8410210c9a

Version string including the ID of the git commit from which it was built, see https://docs.espressif.com/projects/esp ... nt-version

Function ESP.getSdkVersion() returns mess: 1061169513

Unlikely, as it just calls esp_get_idf_version() - and also returns a const char* ("string"), not an integer. You can use macros to evaluate the IDF version: https://docs.espressif.com/projects/esp ... dk-version

Looks like I cannot detect which WiFI version is supported...

I guess you'll want to use esp_chip_info(), and possibly esp_app_get_description().
Btw, the features of your specific target chip are known at compile/build time, so you can check the IDF-defined macros in your code and don't have to call a function at runtime for that.

Thks, it explains. And you are right about ESP.getSdkVersion(), it was my fault

>>I guess you'll want to use esp_chip_info(), and possibly esp_app_get_description().
Thanks for pointing me to this; however, the meaning of
chip_info fields is more or less clear, while the structure returned by the esp_app_get_description() function contains fields whose descriptions are unclear (like "Magic word" or "sha256 of elf file"). Can I clarify it somewhere?

Don't worry about magic word; if it isn't right, you will never boot.
SHA256 is a checksum that the OTA & bootloader use to verify the firmware blob.
The rest of the values are pretty self explanatory, or else constants that will not be very useful to you.