On Thu, 06 May 2010 22:46:42 -0400, Michel Fortin
<michel.for...@michelf.com> wrote:
On 2010-05-06 21:48:09 -0400, "Robert Jacques" <sandf...@jhu.edu> said:
On Thu, 06 May 2010 20:57:07 -0400, Michel Fortin
<michel.for...@michelf.com> wrote:
On 2010-05-06 19:02:03 -0400, Jason House
<jason.james.ho...@gmail.com> said:
Don Wrote:
x[] = sin(y[]);
I strongly favor the first syntax since it matches how I'd write it
in a for loop.
i.e. I'd replace [] with [i].
This is the best way to see array operations I've read up to now:
replace [] with [i], i being the current loop index. It's so simple
to explain.
If there was a sin variant that took array input, then I'd expect
the line to be:
x[] = sin(y)[]
which would translate to creating a temporary to hold sin(y) array.
Makes sense too.
this:
for(int i = 0; i < x.length; i++) {
x[i] = sin(y)[i];
}
makes sense?
Yes, if as stated by Jason there was a sin variant that took array input.
That said, I'd expect the compiler to call sin(y) only once, so it'd be
more like that:
auto sinY = sin(y);
for(int i = 0; i < x.length; i++) {
x[i] = sinY[i];
}
Ah, thank you for the clarification. I mis-read the first post. Although,
for sin(y)[] to work, the variant would have to return an array in
addition to taking one.
hmm...
given
real foo(real value) {...}
and
real[] foo(real[] value) {...}
what should happen with the follow line of code:
x[] = foo(y[]) + z[];
