5 ms·

That's just not true, there are array expressions, and arrays are part of the type system. I give him more credit than that. (The "you can't copy them" is techn

• jstimpfle 6y ago
  You're right, I forgot about compound literals, which were introduced in C99 and which are a rarely used feature. But I guess my statement that "C doesn't have array expressions" was true before the advent of C99. And that's also why array decay made even more sense back then. (It still makes a lot of sense today IMO).
  • hvdijk 6y ago
    I didn't mean compound literals, I do not really see how they change things here, I meant that there are a few cases where arrays don't decay to pointers, and supporting those requires compilers to make arrays a part of the type system. Example: given "int a[3];", how else would you compute (&a+1) ?
    • jstimpfle 6y ago
      To me, the question is what is actually "the type system" and what is "the allocation system" or whatever else aspect of implementing a compiler. So determining the type of "&a" is not an issue, it's just one case in determining the type of a C expression (look up the object "a", is it an array? The type of the expression is a pointer to the array element type). This is not a special case, at least not more special than how to determine the type of the expression "a", or "1". No array type needed.
      • hvdijk 6y ago
        Are you aware that given int a[3];, (&a+1) and (a+3) denote the same address? If you are, how can that possibly work if as you suggest, &a and a are indistinguishable to the compiler, that they are both seen as a pointer to int?
        • jstimpfle 6y ago
          That came quite honestly as a huge surprise to me. I've never seen this before. Thanks for the heads up!