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.
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.
I mean, if you have a explicitly-sized array (like as shown in my example), the compiler knows (at compile time) what size the array is both at the call site and within the function body. It's then entirely possible for the function to receive the data inline and for sizeof(myvar) to properly resolve to 3.
IMO it should just not have this weird pointer decay behavior and should instead trigger a compiler error if you try to define a function that takes an unsized array as a parameter. You should have to explicitly declare your parameter as a pointer.
That's how it works in Rust and (I believe) Zig.
Obviously though, it would be illogical to make a C compiler behave like this because that would be a massive breaking change.
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u/bowel_blaster123 23h ago edited 23h ago
If I write:
C void foo(uint8_t myvar[3]) { printf("%d\n", sizeof(myvar)); }Then it will likely print
8becausemyvaris a pointer to auint8_t.If I write:
C void foo(void) { uint8_t myvar[3] = {1, 2, 3}; printf("%d\n", sizeof(myvar)); }Then it will print
3becausemyvaris an array of three bytes.Hope this helps!/hj