Command line arguments with spaces to a C program through shell-wrapper script

Question

what does it take to make my program accept command-line arguments with spaces?

Yet-another EDIT: I have just recognized that the program is started from a shell-script that sets up the environment for the execution of the program. As there are some external libraries, LD_LIBRARY_PATH is set to the current working directory:

#!/bin/sh

ARCH=`uname -m`
export LD_LIBRARY_PATH=".:lib/magic/linux-${ARCH}"

./data_sniffer.bin $*

exit $?

The issue is definitely related to $*. Is there any solution to correctly forward the command-line parameters?

Here is a code-snippet from main():

if (stat(argv[1], &lStat) != 0)
{   
    fprintf(stderr, "Cannot stat(2) given file: %s. \n", argv[1]);
    return EXIT_FAILURE;
}

I am starting my program with the following parameters:

./data_sniffer /mnt/pod/movies/some\ test\ movie\ file.avi

The resulting error message looks like this: Cannot stat(2) given file: /mnt/pod/movies/some.

Anyone an idea what's wrong here? I think that I am not the first one with this problem (though, I could not find a related question here).

Hi Mat, I am sorry, but this was a mistake. Please think that it is argv[1] (already edited in the question). — dubbaluga
– dubbaluga, Commented Jul 14, 2012 at 13:06
My answer has been updated with relevant information to solve your problem, in the future do not post a question and then decide to change it - it's very annoying to us who decide to answer the first one and then realize that it was completely pointless. — Filip Roséen
– Filip Roséen, Commented Jul 14, 2012 at 13:55
I am sorry for doing this. Maybe I should start a new question instead of changing the existing one the next time. Thank you for your useful answer. It solved my problem. — dubbaluga
– dubbaluga, Commented Jul 14, 2012 at 15:39

Community · Accepted Answer · 2020-06-20 09:12:55Z

8

Replace the use of ./data_sniffer.bin $* with ./data_sniffer.bin "$@" in your wrapper script and the arguments should be forwarded in a correct manner.

More regarding the difference between $*, $@ and "$@" can be found here.

^{the author of this question changed it completely and came forward with more information regarding the matter, I will let everything already written stand but please remember that this was written before his/her last edit..}

Regarding `argv`..

It doesn't require much from you as a developer, I'm tempted (and it's much more truthful) to say that it requires nothing of you.

Arguments passed to the application is handled by the shell executing the binary, doing this below should definitely work the way you want it to (even though I find it odd that you are claiming that the current shell doesn't handle '\ ' correctely):

./data_sniffer '/mnt/pod/movies/some test movie file.avi'
./data_sniffer /mnt/pod/movies/some\ test\ movie\ file.avi

# there should be no difference between the two

My recommendation(s)

Have you tried doing printf ("argv[1]: %s\n", argv[1]); in the beginning of main to validate the contents of it?
Are you sure that you are invoking the correct binary with the correct command-line arguments?

Sadly the only reasonable thing to write is that you are doing something wrong. We are however unable to answer what without further information regarding the issue.

^{I find it very hard to believe that there is a bug in your shell, even though that is of course possible - I doubt it.}

Parsing the command-line into argv isn't something that you as a developer should worry about, there are no enforced functionality that you have to implement for the binary itself to handle spaces in it's argument(s).

edited Jun 20, 2020 at 9:12

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answered Jul 14, 2012 at 12:54

Filip Roséen

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1 Comment

dubbaluga Over a year ago

It does not work for me, both with quotes and backslash-escaping. Actually, that is what I was thinking too: the command-line arguments are put into argv as given: un-escaped (backslashes removed; spaces preserved).

Derui Si · Accepted Answer · 2012-07-14 13:35:19Z

4

In my CentOS 5.3, I made a test.

#include <stdio.h>
int main(int argc, char** argv)
{
        printf("argc=%d\n", argc);
        int i = 0;
        for(i = 0; i<argc; i++)
        {   
                printf("argv[%d]=%s\n", i, argv[i]);
        }   
        return 0;
}

And then run it with different parameters:

[root@dw examples]# ./a.out a b c
argc=4
argv[0]=./a.out
argv[1]=a
argv[2]=b
argv[3]=c
[root@dw examples]# ./a.out 'a b c'
argc=2
argv[0]=./a.out
argv[1]=a b c
[root@dw examples]# ./a.out a\ b\ c
argc=2
argv[0]=./a.out
argv[1]=a b c
[root@dw examples]# ./a.out "a b c"
argc=2
argv[0]=./a.out
argv[1]=a b c

So looks like (1) 'a b c' ; (2) a\ b\ c; (3) "a b c" all are working.

answered Jul 14, 2012 at 13:35

Derui Si

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1 Comment

dubbaluga Over a year ago

Thank you so much for performing a test setting! Please see my new edit. The issue is related to the wrapper script (did not think of it as a source of error) that invokes my program.

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