4 ms·

Since this is about C declarations: for anyone who (like me) had the misfortune of learning the so-called "spiral rule" in college rather than being taught how

by pcfwik 10mo ago

Since this is about C declarations: for anyone who (like me) had the misfortune of learning the so-called "spiral rule" in college rather than being taught how declarations in C work, below are some links that explain the "declaration follows use" idea that (AFAIK) is the true philosophy behind C declaration syntax (and significantly easier to remember/read/write). TL;DR: you declare a variable in C _in exactly the same way you would use it:_ if you know how to use a variable, then you know how to read and write a declaration for it. https://eigenstate.org/notes/c-decl https://eigenstate.org/notes/c-decl https://news.ycombinator.com/item?id=12775966 https://news.ycombinator.com/item?id=12775966

nitrix 10mo ago
That is correct. int x, *p, arr[5], fn(), (*pfn)(); Using x, or dereferencing p, or subscripting the array arr, or declaring a function that can be called with fn, or dereferencing the function pointer pfn then calling it, all these things would produce an int. It's the intended way to read/write declarations/expressions. As a consequence, asterisks ends up placed near the identifiers. The confused ones will think it's a stylistic choice and won't understand any of this.
- saagarjha 10mo ago
  Of course, the correct way to use a function pointer is to call it.
  - nitrix 10mo ago
    Yes, the () operator dereference function pointers automatically for you for convenience. There's also the surprise that you can infinitely dereference function pointers as they just yield you more function pointers.
    - korianders 10mo ago
      One baffling thing I see people do with typedefing function pointers is insisting on adding in the pointer part in the typedef which just complicates and hides things. If you want to typedef a function pointer, make a completely ordinary function declaration, then slap 'typedef' at the beginning, done. This does require you to do "foo_func *f" instead of "foo_func f" when declaring variables, but that is just clearer imo. typedef int foo_func(int); // nice typedef int (*foo_func)(int); // why?
      - cryptonector 9mo ago
        Why do you need the `*` to be part of every variable/member declaration?
- any1 10mo ago
  > It's the intended way to read/write declarations/expressions. As a consequence, asterisks ends up placed near the identifiers. You know you don't always have to use things as they were intended? > The confused ones will think it's a stylistic choice and won't understand any of this. Well, I've written it both ways, and the compiler never seems to mind. :) Maybe I should start putting space on both sides of the asterisk; seems like it would be a good way to annoy even more people.
- pwdisswordfishy 10mo ago
  Blame Stroustrup. https://www.stroustrup.com/bs_faq2.html#whitespace https://www.stroustrup.com/bs_faq2.html#whitespace
userbinator 10mo ago
if you know how to use a variable, then you know how to read and write a declaration for it. In other words, the precedence of operators in a declaration have exactly the same precedence as in its use.

by pcfwik 10mo ago

nitrix 10mo ago

That is correct. int x, *p, arr[5], fn(), (*pfn)(); Using x, or dereferencing p, or subscripting the array arr, or declaring a function that can be called with fn, or dereferencing the function pointer pfn then calling it, all these things would produce an int. It's the intended way to read/write declarations/expressions. As a consequence, asterisks ends up placed near the identifiers. The confused ones will think it's a stylistic choice and won't understand any of this.

saagarjha 10mo ago
Of course, the correct way to use a function pointer is to call it.
- nitrix 10mo ago
  Yes, the () operator dereference function pointers automatically for you for convenience. There's also the surprise that you can infinitely dereference function pointers as they just yield you more function pointers.
  - korianders 10mo ago
    One baffling thing I see people do with typedefing function pointers is insisting on adding in the pointer part in the typedef which just complicates and hides things. If you want to typedef a function pointer, make a completely ordinary function declaration, then slap 'typedef' at the beginning, done. This does require you to do "foo_func *f" instead of "foo_func f" when declaring variables, but that is just clearer imo. typedef int foo_func(int); // nice typedef int (*foo_func)(int); // why?
    - cryptonector 9mo ago
      Why do you need the `*` to be part of every variable/member declaration?
any1 10mo ago
> It's the intended way to read/write declarations/expressions. As a consequence, asterisks ends up placed near the identifiers. You know you don't always have to use things as they were intended? > The confused ones will think it's a stylistic choice and won't understand any of this. Well, I've written it both ways, and the compiler never seems to mind. :) Maybe I should start putting space on both sides of the asterisk; seems like it would be a good way to annoy even more people.
pwdisswordfishy 10mo ago
Blame Stroustrup. https://www.stroustrup.com/bs_faq2.html#whitespace https://www.stroustrup.com/bs_faq2.html#whitespace

userbinator 10mo ago

if you know how to use a variable, then you know how to read and write a declaration for it. In other words, the precedence of operators in a declaration have exactly the same precedence as in its use.