On Oct 21, 2008, at 6:14 PM, Waldemar Horwat wrote:
Maciej Stachowiak wrote:
I don't think you can represent tail position in a switch statement
with your "attribute grammar" notion, but it's clear to me that the
statement immediately before a break statement, or else the last
statement in the last case or default clause, is in tail position.
There is no reason this should be any different than the if/else
case.
Regards,
Maciej
Of course, if you do that then you're back to what Dave was
complaining about: break becomes like return in that it can jump
out of various kinds of places. For example:
switch (x) {
case 1: {
if (x) {
g();
break;
}
h();
break;
}
case 2:
try {
h();
break;
} finally ...
case 3:
...
}
Breaking out of an if seems non-problematic to me, you could imagine
rewriting all such if statements where everything after it is in an
else clause, and then it would be clear that in case 1 both g() and
h() are in tail position (assuming the switch statement itself is).
However nothing inside a "try" block can be in tail potion since there
is implied control context. That's a problem with "try", though, not
with "switch". Obviously wrapping with "switch" cannot turn a non-tail
position into a tail position.
You could probably define a rigorous transform to apply to a swtich()
statement that turns it into a series of if / else clauses (possibly
duplicating later cases if there is no break) and apply the usual if
rule to that transform to get case statements into the attribute
grammar more formally. For example the above could transform as
something like:
if (x == 1) {
if (x) {
g();
} else {
h();
}
} else if (x == 2) {
try {
h();
finally ...
... // duplicate case 3 assuming there was no break
} else if (x == 3) {
...
}
Then it's clear that the first two calls can be in tail position but
the third cannot by the usual rules for if(). It doesn't make sense to
treat switch differently just because the syntax is different.
Regards,
Maciej
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