Re: [TUHS] C declarations.

On Sun, May 14, 2017 at 7:34 PM, Arthur Krewat <krewat(a)kilonet.net> wrote: Yes. It's a pointer. Of course it is! But that doesn't mean that it SHOULD be, which is the point. int array[128]; > Want to reference the array in a memcpy? Do: memcpy(destination, array, > sizeof(array)); > Want to reference an element in the array? array[element #] > Want to pass (a reference to) said array to a function? function(array) > Not sure what the kludge is. The kludge is that it works different than everything else for no really good reason. I think C is more consistent when it comes to arrays than a lot of other > languages. At least with C you KNOW it's a pointer to the original. Sure! But if someone had told you, back when you first learned C, that it was an array and had value semantics, would you have thought THAT was strange? "If you want a pointer to the array, use the address operator on the array and pass that; that'll behave like a 'foo*' pointer for an array of foo. Most people do this for efficiency." The point is that it's a thought exercise. It's not that folks don't understand how C works now or the semantics of array-decay-to-pointer; it's just that folks can also imagine a world where things had been done differently. That world doesn't exist, but we can still imagine it. - Dan C.