Passing quoted arguments to C program in a shell script

Question

I have a C program "main" which gets the following parameters:

"a b c d ..." e f g

That's a total of 4 parameters, because of the quotation. I have a text file which each line has these 4 parameters. I made a shell script to run the C program for each of the parameters:

#!/bin/bash
while read line
do
    ./main "$line"
done < $1

The problem is that the C program is recognizing the first parameter, which is quoted, as several separated parameters, as if the quote was being ignored. Among the many things I've tried, it's worth mentioning that I tried changing each quotation in the file to \" and even remove the quotation from the call (./main $line).

ensc · Accepted Answer · 2013-11-20 00:55:12Z

2

#!/bin/bash
while read line
do
    eval set -- $line
    ./main "$@"
done < $1

answered Nov 20, 2013 at 0:55

ensc

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4 Comments

Dietrich Epp Over a year ago

Why eval? Shouldn't it work fine with just set -- $line?

ensc Over a year ago

parameters seem to be escaped (e.g. "a b c d" e f) so they must be passed through eval. If you do not believe me, just test it ;) echo '"1 2 3 4" a b c' | bash -x /tmp/xxxx.sh

Novice Over a year ago

eval set -- $line works, thanks. But I'm not quite sure why (I'm not that experienced in shell scripting). My explanation would be: set -- $line sets the arguments passed to the shell, for some reason I'm unaware of; then eval evaluates it so that $line can be interpreted accordingly, which is going to take the quote into account. Is that correct? Why does set -- $line sets the arguments passed to the shell, namely, $@? Or it doesn't?

ensc Over a year ago

eval causes it to be interpreted as when you type set -- "1 2 3 4" a b c. Without, it will be interpreted as set -- \"1 2 3 4\" a b c; e.g. the " is taken literally. set sets the positional parameters ($@); the -- is used to avoid interpretation of - in front following args.

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