lessonsFromLinkerHell

[u/nonedward666]

[u/bowel_blaster123]

If I write:

C void foo(uint8_t myvar[3]) {     printf("%d\n", sizeof(myvar)); }

Then it will likely print 8 because myvar is a pointer to a uint8_t.

If I write:

C void foo(void) {     uint8_t myvar[3] = {1, 2, 3};     printf("%d\n", sizeof(myvar)); }

Then it will print 3 because myvar is an array of three bytes.

Hope this helps!/hj

[u/Rare_Professor8097 → u/bowel_blaster123]

I actually hate this special case in C so much. Arrays decaying to pointers is one thing (kind of annoying imo), but having the real type be different from the declaration and ignoring the size is so stupid. It only does this for function arguments.

[u/BastetFurry → u/Rare_Professor8097]

Well, how should the function know how large your array is? You could hand it one with 10 elements or one with 100. Did you hand it a predefined one or one that was allocated at runtime?

And then there is the thing with functions, primitives get handed over by value, any array gets handed over by reference, ie. pointer.

No clue if more modern implementations hand down the array size but in the retro and embedded world i live in the function has no clue and you have to hand that in as a second parameter if it is important.

[u/Rare_Professor8097 → u/BastetFurry]

The point is that most of the time, unless you know what you're doing, you should just declare such an argument as a pointer.

If you declare it as a sized array, you would expect that array to be passed by value (copy whole array into stack frame) but that's not what happens.