bearophile wrote:
Paul D. Anderson:
From the D2.0/Language/Functions page, under Function Parameters:
"The in storage class is equivalent to const scope."
That's why the "const in" combination doesn't work.
Why is this working then, and printing 100?
import std.stdio: writeln;
void foo(const ref int x) {
writeln(x * 10);
}
You are not changig x here. The x*10 just multiplies and returns its
result to writeln.
void main() {
int y = 10;
foo(y);
}
Note that the following doesn't compile:
void foo(in ref int x) {
Bye,
bearophile
