On Monday, 26 June 2017 at 10:34:22 UTC, ag0aep6g wrote:
On 06/26/2017 11:51 AM, helxi wrote:
[...]
`a` is a tuple of the run-time arguments you pass to
`sequence`. In this example, no arguments are passed (empty
parens at the end of the call), so `a` is empty.
[...]
a[0] = 1
a[1] = 2
[...]
`a` isn't used in the string lambda, and it's not considered an
element of the range. So n starts at 0 and this just prints
0+2, 1+2, 2+2, etc.
[...]
`a` is still not used in the string lambda, but `recurrence`
uses the values in `a` as the first elements of the range. `n`
is incremented accordingly (to 1), so this prints:
1 = a[0],
3 = (n = 1) + 2,
4 = (n = 2) + 2,
etc.
Another difference between `sequence` and `recurrence` is that
`a` always refers to the same initial value(s) in `sequence`,
while in `recurrence` it gets updated and refers to the
previous element(s) of the range:
----
sequence!((a, n) => a[0] + 1)(1).take(10).writeln;
// [2, 2, 2, 2, 2, 2, 2, 2, 2, 2]
// because a[0] is always 1
recurrence!((a, n) => a[0] + 1)(1).take(10).writeln;
// [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
// because a[0] refers to the previous value
----
Oh thank you. Just 2 follow-up questions:
recurrence!((a, n) => a[0] + 1)(1).take(10).writeln;
1. In the last example of reccurence, what does n in (a,n) refer
to?
2. How would you chain until! with reccurence? For example I want
to compute 1, 10, 100, ..., (until the value remains smaller than
1000_000)?
