Grep and ignoring leading whitespace

Question

I've been working on a bash script to search for names of defines, and then grab the hex values for them and put them in the list. Once I have the list of names I'll attempt to search for "#define [name]" using -w to ensure exact match, and then awk '{ print $3 }' to grab the hex value.

However it works if the line in the header file is similar to

a.h:#define [name] 0x0001

But it does NOT work if it is similar to

a.h:    #define [name] 0x0001

How can I get around this? I have tried this

grep -nrw "\s*#define[[:space:]]*$p" . --include=*.h | awk '{ print $3 }'

I thought the \s* would ignore the leading whitespace before #define but it doesn't. Am I doing something wrong?

Answer 1

Just use awk (using grep seems redundant to me, since awk already can match a regular expression):

awk '$0~/\s*\#define\s*\[.*\]\s*.*/ {print $3}' *.h

Going through the expression in more detail:

$0 ~ /regexp/          # look for the regular expression in the record
      \s*              # whitespace, any number of times
         \#define      # literal string, '#' has to be scaped
                 \s*   # same as above
                    .* # any character, any number of times, this is
                       # your hex code and you can refine the regex here
{ print $3 }           # print the third field if the record matches

To have this run recursively, e.g.

mkdir -p a/b/c
echo "   #define [name] 0x0001" > a/a.h
echo "   #define [name] 0x0002" > a/b/b.h
echo "   #define [name] 0x0003" > a/b/c/c.h
tree
.
└── a
    ├── a.h
    └── b
        ├── b.h
        └── c
            └── c.h

3 directories, 3 files

since awk needs to be given a list of files to operate on, you could:

find . -type f -name "*.h" \
  -exec awk '$0~/\s*\#define\s*\[.*\]\s*.*/ {print $3}' {} \;
0x0002
0x0003
0x0001

Answer 2

Use grep -o to print only the matching part of the line.

Obviously take the \s part out again, because you don't want that part.