On Monday, 1 August 2022 at 14:46:33 UTC, ag0aep6g wrote:
On Monday, 1 August 2022 at 14:39:17 UTC, pascal111 wrote:
On Monday, 1 August 2022 at 14:34:45 UTC, ag0aep6g wrote:
[...]
`a => a > 0` is not a statement. It's an expression.
But it is still a "function", and functions make statements!!
It's not a normal expression.
It's a normal expression.
`foo => bar` is an expression that doesn't involve any
statement. So there's no semicolon.
`(foo) { return bar; }` does contain a return statement. As you
expect, there's a semicolon. But it's still an expression like
any other.
If `foo => bar` == `(foo) { return bar; }`, then `foo => bar` is
a function. "=>" is not an operator, it's a special symbol for
lambda "function".
If A == B, so A's types is the same of B's type. How can it be
withstanding `foo => bar` == `foo => bar` == `(foo) { return bar;
}` and `foo => bar` is an expression and the other is a
function?!! no sense.
