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  • ezchili@iusearchlinux.fyi
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    10 months ago

    There’s a syntax for indexing starting from 0, it’s

    *(&arr+0) to *(&arr+(n-1))

    For the rest of us who are manipulating sets of values and not offsets on pointers and aren’t delusionally attached to conventions, there’s arr[1] to arr[n]

    • barsoap@lemm.ee
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      10 months ago
      ptr[n] == n[ptr] == *(ptr+n) == *(n+ptr).
      

      Addition is commutative so of course array indexing is and why the hell are you taking the address of a pointer. Also it’s not “int pointer foo” but “foo, dereferenced, is an int” that’s why it’s int *foo not int* foo. I won’t die on that mountain fortress because it is unassailable. Never write char **argv (but char *argv[]) but it’s vital to understand why it doesn’t make a difference to the compiler. It’s what passes as self-documenting code in C land.

      Also 0-based indexing is older than C. It’s older than assembly.

      • ezchili@iusearchlinux.fyi
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        10 months ago

        Why do you assume it was a pointer type? There’s no types. Why do you assume C either? This is pseudo code to illustrate pointer offsets

        • barsoap@lemm.ee
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          10 months ago

          Why do you assume it was a pointer type?

          Because afterwards you said arr[n]. By convention n is definitely an integer and if arr is also, say, an integer, you get

           error: subscripted value is neither array nor pointer nor vector
          

          Why do you assume C either?

          Because you didn’t write ^(@arr+0) (Not sure that’s even valid though my Pascal is very rusty).

          This is pseudo code to illustrate pointer offsets

          Granted. But then it’s still Pseudo-C, not Pseudo-Pascal or Pseudo-Whitespace.

          • ezchili@iusearchlinux.fyi
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            10 months ago

            It’s pseudo-nothing

            It conveys a point, which you got, and if you decide to invent a syntax and bicker on it it’s just you

            Really pointless discussion