On Friday, 25 December 2015 at 14:55:04 UTC, anonymous wrote:
On 25.12.2015 13:10, Joakim Brännström wrote:
[B]
Evaluates to the function type "constructed" by binaryFun.
Almost. It evaluates to the type of the expression. The
expression is a function call, so typeof evaluates to the
return type of the generated function.
Ahh, I missed this.
The subtle difference between:
int fun(uint x) { return 1; }
pragma(msg, typeof(fun)); // -> int(uint x)
pragma(msg, typeof(fun()); // -> int
That second one leads us to the longest form of the is-typeof
idiom: `is(typeof({foo(); bar();}))`. Wrapping the code in a
delegate so that it's an expression, which can be passed to
typeof.
Nice, didn't know that was possible. I'll remember that.
Thank you for the detailed answer.
