Align trailing backslashes vertically - awk

Question

I have a file that looks like this:
abc \
d \
efgh \
i
jklmnop \
q \rst \
uv
wx
y \
z
For each group of consecutive lines that have a backslash at the end, I want to arrange those backslashes in a straight vertical line. So the expected output for above sample is:
abc \
d \
efgh \
i
jklmnop \
q \rst \
uv
wx
y \
z
I managed to align all backslashes with this program:
$ awk '/\\$/ { sub(/\\$/,""); printf "%-20s\\\n",$0; next} 1' file
abc \
d \
efgh \
i
jklmnop \
q \rst \
uv
wx
y \
z
But I have no idea how to proceed from here, so I'm asking for guidance. I tried searching on SO, but top results were all about removing trailing backslashes.
Details about the actual input:
Lines may contain any character including backslash and consist of any number of characters, there is no limit.
There might be multiple blanks and tabs before and after the last backslash.
There is always at least one blank or tab preceding the last backslash.
There is no line that consists of only a backslash and zero or more blanks/tabs around it.
P.S. I'm not looking for a Perl solution.

Answer 1

If you don't have any tabs before the spaces-then-backslash-then-spaces at the end of each line:
$ cat tst.awk
FNR == 1 { prevHasEsc = blockNr = 0 }
hasEsc = match($0,/[[:space:]]*\\[[:space:]]*$/) {
$0 = substr($0,1,RSTART-1)
if ( ! prevHasEsc ) {
++blockNr
}
}
{ prevHasEsc = hasEsc }
NR == FNR {
if ( hasEsc ) {
lgth[blockNr] = (lgth[blockNr] > length($0) ? lgth[blockNr] : length($0))
}
next
}
hasEsc {
$0 = sprintf("%-*s \\", lgth[blockNr], $0)
}
{ print }
.
$ awk -f tst.awk file file
abc \
d \
efgh \
i
jklmnop \
q \
r
If you do then I'd suggest running the text through pr -e -t first to convert tabs to the corresponding number of blanks.

Answer 2

Here is one attempt using gnu-awk with a custom RS that breaks input on each substring that ends without a blackslash:
awk -v RS='[^\n]*[^\\\\[:space:]][[:blank:]]*(\n|$)' '
{
sub(/\n$/, "", RT)
}
n = split($0, lines, /[[:blank:]]+\\[[:blank:]]*\n/) {
lines[n] = RT
mlen = 0
# determine max length of a block
for (i=1; i<=n; i++)
if (mlen < length(lines[i]))
mlen = length(lines[i])
# print each segment with backslash at the end of max length
for (i=1; i<n; i++)
printf "%-" (mlen+1) "s\\\n", lines[i]
}
RT {
print RT
}' file
abc \
d \
efgh \
i
jklmnop \
q \rst \
uv
wx
y \
z
Code Demo
Details:
-v RS='[^\n]*[^\\\\[:space:]][[:blank:]]*(\n|$): Sets record separator using this regex which basically matches a line that doesn't end with a \. As a result we'll get all contiguous lines that end with \ in each record.
split($0, lines, /[[:blank:]]+\\[[:blank:]]*\n/: Splits each record by ending \ and following line break.
In the first for loop, we loop through each array element and determine longest length of the line i.e. mlen
In the second for loop, we just print each line segment using mlen+1 as length to place trailing \
Finally we print RT which is the substring we capture as a result of RS=...