ignore pattern with backward slash in diff command

Question

File foo.c:

do {                                                            
    a=b;
    c=d;
 } while (0)

File bar.c

do {                                                            
    a=b;
    c=d;
\
} while (0)

In shell script the diff of above 2 files:

diff foo.c bar.c

<   } while (0)
---
>     \
>   } while (0)

I am trying to ignore this diff using

diff -I "reg-ex pattern".

I tried pattern like "\s}\swhile\s(0)" , "} while (0)" etc. But no use. Is it possible to ignore the above case? If not "diff", any other utility?

Answer 1

You can do that by deleting the unwanted lines with sed. And then ignoring the differences.

You have 2 differences between foo.c and bar.c

1. Line with \
2. There is a space before } in foo.c

You can do something like this:

diff -I '\s\?}' <(sed '/^\\/d' bar.c) foo.c

Delete the line with \ using sed and ignore the space in foo.c

Answer 2

Try this:

pat='^[[:space:]]*\\[[:space:]]*$'
sub_pat='s/'"$pat"'//'
diff -B -I "$pat" <(sed -e "$sub_pat" foo.c) <(sed -e "$sub_pat" bar.c)

The first line stores a regular expression into pat variable for the next diff and sed commands. The regular expression matches a line with a backslash (\\) optionally surrounded with space characters.

The sed commands replace the lines matching $pat pattern with empty strings (-B option causes diff to ignore blank lines).

---

It is impossible to filter out the backslashes from the output of diff without extra tools, unless the only difference is in the lines containing the pattern being ignored:

for each nonignorable change, `diff' prints the complete set of changes in its vicinity, including the ignorable ones.

See answers to this question for details.