[Answered]: Musing on ulp_load_binary() samples ...

[kolban]

Am studying the ULP co-processor and looking at samples. I came across this one from our examples repository:

https://github.com/espressif/esp-idf/bl ... main.c#L45

The code I am looking at reads as follows:

Code: Select all

ulp_load_binary(0, ulp_main_bin_start, (ulp_main_bin_end - ulp_main_bin_start) / sizeof(uint32_t));

Now imagine the binary is 40 bytes long ... that would give us 40 div 4 = 10 32bit words as the size parameter. Great. But now imagine that the binary is 41 bytes long. 41 div 4 is still 10 ... and we have lost a byte.

If the assembler assures that its binary result is always 32bit aligned ... then no issues and end of story. Which means that my question boils down to "Is the output of an assembly of a ULP co-processor source always going to be evenly divisible by 4?"

[ESP_igrr]

Each ULP instruction is 32-bit, so assembler output size will be divisible by 4.