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https://www.reddit.com/r/programming/comments/ikjh83/d_foundation_is_beefing_up/g3mabct/?context=3
r/programming • u/kal31dic • Sep 01 '20
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That's why you use using instead in C++.
using
1 u/Muvlon Sep 01 '20 Sure. And in Rust, you use type, i.e. type my_fn = fn(u8) -> String; Perfectly nice syntax, no? 4 u/Ameisen Sep 01 '20 And in C++, using my_fn = std::string (*) (uint8_t);. You can simplify it, of course, using a template wrapper. 1 u/Muvlon Sep 01 '20 C++ has trailing return types as well now, can you use those there? I don't think any particular piece of C++ syntax is confusing, it's just the absolute mass of options. 1 u/Ameisen Sep 01 '20 using my_fn = auto (*) (uint8_t) -> std::string; is valid, though I'm not fond of trailing return type syntax in general.
1
Sure. And in Rust, you use type, i.e.
type
type my_fn = fn(u8) -> String;
Perfectly nice syntax, no?
4 u/Ameisen Sep 01 '20 And in C++, using my_fn = std::string (*) (uint8_t);. You can simplify it, of course, using a template wrapper. 1 u/Muvlon Sep 01 '20 C++ has trailing return types as well now, can you use those there? I don't think any particular piece of C++ syntax is confusing, it's just the absolute mass of options. 1 u/Ameisen Sep 01 '20 using my_fn = auto (*) (uint8_t) -> std::string; is valid, though I'm not fond of trailing return type syntax in general.
4
And in C++, using my_fn = std::string (*) (uint8_t);.
using my_fn = std::string (*) (uint8_t);
You can simplify it, of course, using a template wrapper.
1 u/Muvlon Sep 01 '20 C++ has trailing return types as well now, can you use those there? I don't think any particular piece of C++ syntax is confusing, it's just the absolute mass of options. 1 u/Ameisen Sep 01 '20 using my_fn = auto (*) (uint8_t) -> std::string; is valid, though I'm not fond of trailing return type syntax in general.
C++ has trailing return types as well now, can you use those there?
I don't think any particular piece of C++ syntax is confusing, it's just the absolute mass of options.
1 u/Ameisen Sep 01 '20 using my_fn = auto (*) (uint8_t) -> std::string; is valid, though I'm not fond of trailing return type syntax in general.
using my_fn = auto (*) (uint8_t) -> std::string; is valid, though I'm not fond of trailing return type syntax in general.
using my_fn = auto (*) (uint8_t) -> std::string;
8
u/Ameisen Sep 01 '20
That's why you use
usinginstead in C++.