Parsing each field and process it using 'awk'/'gawk'

Question

Hi all,

Here is a query:

grep bar 'foo.txt' | awk '{print $3}'

The field name emitted by the 'awk' query are mangled C++ symbol names. I want to pass each to dem and finally output the output of 'dem'- i.e the demangled symbols.

Assume that the field separator is a ' ' (space).

Answer 1

How about

grep bar 'foo.txt' | awk '{ print $3 }' | xargs dem | awk '{ print $3 }'

Answer 2

awk is a pattern matching language. The grep is totally unnecessary.

awk '/bar/{print $3}' foot.txt

does what your example does.

Edit Fixed up a bit after reading the comments on the precedeing answer (I don't know a thing about dem...):

You can make use of the system call in awk with something like:

awk '/bar/{cline="dem " $3; system(cline)}' foot.txt

but this would spawn an instance of dem for each symbol processed. Very inefficient.

So lets get more clever:

awk '/bar/{list = list " " $3;}END{cline="dem " list; system(cline)}' foot.txt

BTW-- Untested as I don't have dem or your input.

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Another thought: if you're going to use the xargs formulation offered by other posters, cut might well be more efficient than awk. At that point, however, you would need grep again.

Answer 3

This will print the demangled symbols, complete with argument lists in the case of methods:

awk '/bar/ { print $3 }' foo.txt | xargs dem | sed -e 's:.* == ::'

This will print the demangled symbols, without argument lists in the case of methods:

awk '/bar/ { print $3 }' foo.txt | xargs dem | sed -e 's:.* == \([^(]*\).*:\1:'

Cheers, V.