Re: Compile time foreach with switch
On Saturday, 22 April 2017 at 11:56:31 UTC, Nick Treleaven wrote: If I compile the above I get: Deprecation: switch case fallthrough - use 'goto default;' if intended IMO the compiler should issue this warning with the OP's code. https://issues.dlang.org/show_bug.cgi?id=7390#c4
Re: Compile time foreach with switch
On Friday, 21 April 2017 at 19:17:32 UTC, Adam D. Ruppe wrote:
Then realize that it does exactly the same thing when it is
inside a switch... it always breaks the innermost thing, which
happens to be this loop. (note that cases are not `breaked`,
the switch is.)
By coincidence I ran into this issue recently and assumed that
break should break the switch too (and I hadn't read the
boost::hana thread). Thanks for explaining, I thought it was
probably a compiler bug and hadn't gotten round to investigating
further.
So the compiler generates something like this:
int s = 0;
switch (s) {
case 0:
writeln("matched ");
default:
writeln("no match");
break;
}
If I compile the above I get:
Deprecation: switch case fallthrough - use 'goto default;' if
intended
IMO the compiler should issue this warning with the OP's code.
Re: Compile time foreach with switch
On Fri, Apr 21, 2017 at 07:17:32PM +, Adam D. Ruppe via Digitalmars-d-learn
wrote:
> On Friday, 21 April 2017 at 19:09:25 UTC, Johan Fjeldtvedt wrote:
> > void foo(string s) {
> > enum es = tuple("a", "b", "c");
> > switch (s) {
> > foreach (e; es) {
> > case e:
> > writeln("matched ", e);
> > break;
> > }
>
> Let me remove some surrounding stuff and ask you what happens:
>
>foreach (e; es) {
> if(s == c) {
> writeln("matched ", e);
> break;
> }
>}
>
>
> What does that break do?
>
>
> Then realize that it does exactly the same thing when it is inside a
> switch... it always breaks the innermost thing, which happens to be
> this loop. (note that cases are not `breaked`, the switch is.)
Yeah, you want to label your switch statement, say `mySwitch:`, so that
you can do a `break mySwitch;`
T
--
Knowledge is that area of ignorance that we arrange and classify. -- Ambrose
Bierce
Re: Compile time foreach with switch
On Friday, 21 April 2017 at 19:17:32 UTC, Adam D. Ruppe wrote:
On Friday, 21 April 2017 at 19:09:25 UTC, Johan Fjeldtvedt
wrote:
void foo(string s) {
enum es = tuple("a", "b", "c");
switch (s) {
foreach (e; es) {
case e:
writeln("matched ", e);
break;
}
Let me remove some surrounding stuff and ask you what happens:
foreach (e; es) {
if(s == c) {
writeln("matched ", e);
break;
}
}
What does that break do?
Then realize that it does exactly the same thing when it is
inside a switch... it always breaks the innermost thing, which
happens to be this loop. (note that cases are not `breaked`,
the switch is.)
*facepalm* Of course! Thanks.
Re: Compile time foreach with switch
On Friday, 21 April 2017 at 19:09:25 UTC, Johan Fjeldtvedt wrote:
void foo(string s) {
enum es = tuple("a", "b", "c");
switch (s) {
foreach (e; es) {
case e:
writeln("matched ", e);
break;
}
Let me remove some surrounding stuff and ask you what happens:
foreach (e; es) {
if(s == c) {
writeln("matched ", e);
break;
}
}
What does that break do?
Then realize that it does exactly the same thing when it is
inside a switch... it always breaks the innermost thing, which
happens to be this loop. (note that cases are not `breaked`, the
switch is.)
Compile time foreach with switch
I was a bit surprised to find out
(https://forum.dlang.org/post/csiwyetjkttlxxnwn...@forum.dlang.org) that compile time foreach-loops can be used inside switch-statements. I tried the following:
import std.stdio;
import std.typecons;
void foo(string s) {
enum es = tuple("a", "b", "c");
switch (s) {
foreach (e; es) {
case e:
writeln("matched ", e);
break;
}
default:
writeln("no match");
break;
}
}
void main() {
foo("a");
}
However, this prints both "matched a" and "no match". It seems
like either the break or the default: label is ignored. In this
example I could have used return of course, but is this behavior
correct?
