When parsing a time only string like:

```
datetime.datetime.strptime('13:48:25', '%H:%M:%S')
```

This produces:

```
datetime.datetime(1900, 1, 1, 13, 48, 25)
```

Not that the year is `1900` which just doesn't make sense. This will produce `-1` when you attempt to get it's UNIX timestamp. And while that sounds weird it is very useful.

The default year should be 1970. And this will resolve the issue.

The year 1900 is a delibrate choice, it's even documented:
https://docs.python.org/dev/library/datetime.html#strftime-and-strptime-behavior

"For time objects, the format codes for year, month, and day should not be used, as time objects have no such values. If they’re used anyway, 1900 is substituted for the year, and 1 for the month and day."

> The default year should be <whatever>

If you expect a year, you can easily use dt.replace(year=<whatever>), no?

Or start with a datetime.time object and then build a datetime.datetime using the time? It's up to you.

We will not change a default value, it will break the backward compatibility.