On Friday, 11 May 2012 at 23:07:31 UTC, Vidar Wahlberg wrote:
Thank you for the detailed answer.
I still suspect this can fool some people, and (in my
ignorance) I couldn't (and still can't, to be honest) really
see when you would want to assign a variable in a lazy
parameter, I would expect t
On 2012-05-12 00:27, Chris Cain wrote:
[...]
Thank you for the detailed answer.
I still suspect this can fool some people, and (in my ignorance) I
couldn't (and still can't, to be honest) really see when you would want
to assign a variable in a lazy parameter, I would expect that to be far
m
On Friday, 11 May 2012 at 21:39:57 UTC, Vidar Wahlberg wrote:
I'm not suggesting that the compiler should print a warning if
you're doing a calculation in the function call, I'm suggesting
it should give you a warning if you're assigning the result of
the calculation to a variable in the functi
On Friday, May 11, 2012 23:39:44 Vidar Wahlberg wrote:
> On 2012-05-11 23:03, Chris Cain wrote:
> > On Friday, 11 May 2012 at 20:45:53 UTC, Vidar Wahlberg wrote:
> >> Perhaps the compiler should print out a warning when you're assigning
> >> a value to a lazy parameter in a function call?
> >
> >
On 2012-05-11 23:03, Chris Cain wrote:
On Friday, 11 May 2012 at 20:45:53 UTC, Vidar Wahlberg wrote:
Perhaps the compiler should print out a warning when you're assigning
a value to a lazy parameter in a function call?
The entire point of a lazy parameter is to not be
calculated/processed unti
On Friday, 11 May 2012 at 20:45:53 UTC, Vidar Wahlberg wrote:
I often call functions where one of the parameters may be an
integer which i post/pre increment/decrement. However, that can
be quite risky if the parameter is defined as "lazy" as shown
above.
The value of "a" above after calling "l
Wasn't easy to find a short good description of the issue in the
subject, but here's some code to illustrate my "concern":
---
import std.stdio;
void log(T...)(lazy string message, lazy T t) {
debug writefln(message, t);
}
void main() {
int a = 42;
writefln("The meaning of
