On Friday, 27 May 2016 at 08:59:43 UTC, Marc Schütz wrote:
Yes indeed it does. Thanks. Something in my version must have
been different.
On Thursday, 26 May 2016 at 10:51:30 UTC, John Nixon wrote:
On Wednesday, 25 May 2016 at 15:44:34 UTC, Marc Schütz wrote:
On Tuesday, 24 May 2016 at 20:58:11 UTC, John Nixon wrote:
On Tuesday, 24 May 2016 at 15:17:37 UTC, Adam D. Ruppe wrote:
On Tuesday, 24 May 2016 at 14:29:53 UTC, John
On Wednesday, 25 May 2016 at 15:44:34 UTC, Marc Schütz wrote:
On Tuesday, 24 May 2016 at 20:58:11 UTC, John Nixon wrote:
On Tuesday, 24 May 2016 at 15:17:37 UTC, Adam D. Ruppe wrote:
On Tuesday, 24 May 2016 at 14:29:53 UTC, John Nixon wrote:
Or add an explicit constructor:
struct CS {
On Tuesday, 24 May 2016 at 20:58:11 UTC, John Nixon wrote:
On Tuesday, 24 May 2016 at 15:17:37 UTC, Adam D. Ruppe wrote:
On Tuesday, 24 May 2016 at 14:29:53 UTC, John Nixon wrote:
This naively doesn’t seem right because the RHS of an
assignment should not be altered by it.
It's because the
On Tuesday, 24 May 2016 at 15:17:37 UTC, Adam D. Ruppe wrote:
On Tuesday, 24 May 2016 at 14:29:53 UTC, John Nixon wrote:
This naively doesn’t seem right because the RHS of an
assignment should not be altered by it.
It's because the char[] being shallow copied still leads to
mutable stuff.
On Tuesday, 24 May 2016 at 14:29:53 UTC, John Nixon wrote:
This naively doesn’t seem right because the RHS of an
assignment should not be altered by it.
It's because the char[] being shallow copied still leads to
mutable stuff.
What I typically do here is just add a method `dup` to the
1 import std.stdio;
2
3 struct CS{
4 char[] t;
5 CS opAssign(const CS rhs){
6 writeln("CS.opAssign called");
7 this.t = rhs.t.dup;
8 return this;}
9 };
10 void test_fun(const ref CS rhs){
11 CS cs = rhs;//error cannot implicitly convert expression
(rhs) of
