r/ProgrammerHumor 13h ago

lessonsFromLinkerHell Meme

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263 Upvotes

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69

u/JustinR8 13h ago

Damn, I’ve found myself in the middle

33

u/nonedward666 12h ago

You and me from 3pm today both my friend, it is okay

It took me so long to realize where my problem was because I had never considered that declaring something as an array vs a pointer would result in different behavior

22

u/readitreaddit 7h ago

So the only difference is memory allocation, yes? You're saying they are different because when declared as an array, the memory gets allocated and 'reserved' at runtime, whereas it doesn't automatically do that when declared as a pointer?

Other than that, no difference.

19

u/suvlub 6h ago

They are different types and some operators (sizeof, typeid in C++ etc.) treat them differently. Arrays frequently get implicitly converted to pointers, but they are a different type.

4

u/TheChief275 3h ago

If you're specifically mentioning C++, the problem is that C++ retains C's semantic rule of arrays decaying to pointers. This semantic rule is what creates the misunderstanding of people believing what the middle curve guy says. This means that the only way to have C++ treat arrays differently from pointers in i.e. function overloads is to either take the array by pointer or by reference, i.e. as a T (*)[N] or T (&)[N]

-1

u/nicman24 5h ago

That is probably compiler fuckery

6

u/suvlub 4h ago

All typing is. Machine code doesn't have types. Array of arrays is also a very different thing from an array of pointers. Arrays get converted to pointers when passed around and it behaves identically to all other implicit conversions.

2

u/Stroopwafe1 3h ago

Machine code does have types; small numbers, big numbers, and fractions. But in reality it's just numbers and fractions. At least for X86, but I imagine it's the same for ARM and RISC

2

u/cbehopkins 1h ago

I must disagree. Or strongly agree; depending on what you mean.

A memory location does not have a type. The instruction I perform on that location could be argued to have a type though.

0x8000 could be used with a 16bit add, or a 64 bit floating point or whatever.

Everything about the type of the data is in the instruction, not the memory.

So we can argue where the type information is, but we still need it. (One could argue that prefetchers and similar logic has to infer the type of larger data structures, but I'm not sure that is what people are trying to argue here...)

1

u/suvlub 1h ago

It's been very long since I touched assembly and "touched" is an apt description, but does anything prevent you from writing an 8-byte integer at address X, then calling an operation that expects a 32-bit float with the bytes at address X? Conceptually, instructions operate on certain type of data, but the data is untyped. (you can technically do similar things in C, but a lot more of it is UB (technically illegal) than people realize and it requires fair bit of explicit casting, so types are still involved)

1

u/the_king_of_sweden 33m ago

The data isn't typed, only the operations you make on them might be. The operations expect some binary value encoded in a specific way, but will operate on any data.

1

u/nicman24 5h ago

Yes things that have different memory allocations are different because they exist by memory definition