Zero Initialize a Type

Question

Given a variable of an unsigned integral type: foo lets say I want to do this:

const decltype<foo> bar{};

cout << (55834574890LL & ~bar) << endl;

That gives me the expected 42. But now let's say that I want to do away with the bar variable. So something like this:

cout << (55834574890LL & ~decltype<foo>{}) << endl;

But I just get an error:

error: expected primary-expression before decltype

I've also tried declval but that returns a reference, which is also no good. Is there a way I can do this?

`decltype bar{}` is this syntax valid? You probably meant `decltype(foo) bar{}` — lakshayg, Jun 01 '18 at 21:31
What compiler are you using? `const decltype bar{};` should not compile: http://coliru.stacked-crooked.com/a/48d19c6ef9ec5907 — NathanOliver, Jun 01 '18 at 21:31
Was it a good idea to edit the original mistake out of the question? Surely this will just confuse anyone reading it for the first time. — Paul Sanders, Jun 02 '18 at 22:10
@PaulSanders Yeah I guess you're right I've rolled that back. — Jonathan Mee, Jun 04 '18 at 12:06

score 2 · Accepted Answer · answered Jun 01 '18 at 21:30

2

You should use round brackets:

auto v = 55834574890LL & ~decltype(foo){};

answered Jun 01 '18 at 21:30

joe_chip

1 Answers1