On Sep 30, 2019, at 4:45 AM, Jim Laskey <james.las...@oracle.com> wrote:
>
> During the discussion on Text Blocks, several of you stated a need for a line
> continuation construct. I have since created a CSR to propose the creation of
> two new escape sequences: \<line terminator> and \s.
>
> https://bugs.openjdk.java.net/browse/JDK-8227870
>
One observation: Some traditional uses of \LT, as in C macros, makefiles, and
shell,
line up the \ characters in a single column as a sort of right-hand fence:
foo bar \
baz \
bat
This can be done only in settings where the spaces before the \ are treated as
ancillary format, not payload characters. We could make a similar rule for
multi-line
literals, by saying that unescaped spaces *before* the \LT are also deleted when
the \LT sequence is deleted.
In other words, the \ at the end of the line is *not* a fence that transforms
the
So:
var x = “””
foo bar \
baz \
bat \
“””;
assert x.equals(“foo bar baz bat”);
The single space before baz and the three spaces before bat are the left-hand
leading spaces, not the ancillary spaces before the \ characters.
If some of those trailing spaces are desirable, then \s can be used to make
a fence that saves them from stripping.
var x = “””
foo bar\s \
baz \s \
bat\s \
“””;
assert x.equals(“foo bar baz bat ”);
This rule decouples the two functions of \LT in the current proposal: (1) It
joins
lines, and (2) it creates a fence which makes previously ancillary spaces into
real payload spaces. These are distinct jobs and should not be conflated.
I think it’s a small but real improvement to separate the jobs, and allow \s
to handle (2) and \LT to handle only (1).
— John
P.S. A variation of the above suggestion, probably less preferable, would
delete
spaces both *before* and *after* \LT, leading to this:
var x = “””
foo bar \
baz \
bat \
“””;
assert x.equals(“foo barbazbat”);
Again, \s could save spaces from stripping.
