This is a duration method for the `seq()`

generic.

Using `seq()`

on duration objects always retains the type of `from`

.

When calling `seq()`

, exactly two of the following must be specified:

`to`

`by`

Either

`length.out`

or`along.with`

## Usage

```
# S3 method for class 'clock_duration'
seq(from, to = NULL, by = NULL, length.out = NULL, along.with = NULL, ...)
```

## Arguments

- from
`[clock_duration(1)]`

A duration to start the sequence from.

- to
`[clock_duration(1) / NULL]`

A duration to stop the sequence at.

`to`

is cast to the type of`from`

.`to`

is only included in the result if the resulting sequence divides the distance between`from`

and`to`

exactly.- by
`[integer(1) / clock_duration(1) / NULL]`

The unit to increment the sequence by.

If

`by`

is an integer, it is transformed into a duration with the precision of`from`

.If

`by`

is a duration, it is cast to the type of`from`

.- length.out
`[positive integer(1) / NULL]`

The length of the resulting sequence.

If specified,

`along.with`

must be`NULL`

.- along.with
`[vector / NULL]`

A vector who's length determines the length of the resulting sequence.

Equivalent to

`length.out = vec_size(along.with)`

.If specified,

`length.out`

must be`NULL`

.- ...
These dots are for future extensions and must be empty.

## Details

If `from > to`

and `by > 0`

, then the result will be length 0. This matches
the behavior of `rlang::seq2()`

, and results in nicer theoretical
properties when compared with throwing an error. Similarly, if `from < to`

and `by < 0`

, then the result will also be length 0.

## Examples

```
seq(duration_days(0), duration_days(100), by = 5)
#> <duration<day>[21]>
#> [1] 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80
#> [18] 85 90 95 100
# Using a duration `by`. Note that `by` is cast to the type of `from`.
seq(duration_days(0), duration_days(100), by = duration_weeks(1))
#> <duration<day>[15]>
#> [1] 0 7 14 21 28 35 42 49 56 63 70 77 84 91 98
# `to` is cast from 5 years to 60 months
# `by` is cast from 1 quarter to 4 months
seq(duration_months(0), duration_years(5), by = duration_quarters(1))
#> <duration<month>[21]>
#> [1] 0 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60
seq(duration_days(20), by = 2, length.out = 5)
#> <duration<day>[5]>
#> [1] 20 22 24 26 28
```