How to run a c program in bash script and give it 2 arguments?

Question

I have a program in C, which takes 2 arguments, filename and text. I want to write a script in bash, which also take 2 arguments, path and file extension, will iterate through all files in given path and give to my program in C as argument files with the givenextension only and text.

Heres my program in C, nothing special really:

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char **argv)
{
    if(argc < 3)
    {
        fprintf(stderr, "Give 2 args!\n");
        exit(-1);
    }

    char *arg1 = argv[1];
    char *arg2 = argv[2];

    fprintf(stdout, "You gave: %s, %s\n", arg1, arg2);

    return 0;
}

and my bash script:

#!/bin/bash

path=$1
ext=$2
text=$3

for file in $path/*.$ext
do
    ./app | 
    {
        echo $file 
        echo $text
    }
done

I use it like this: ./script /tmp txt hello and it should give as arguments all txt files from /tmp and 'hello' as text to my C program. No it only shows Give 2 args! :( Please, help.

Some programmer dude · Accepted Answer · 2013-01-30 10:15:51Z

13

Right now you're not actually passing any arguments to the program in your script.

Just pass the arguments normally:

./app "$file" "$text"

I put the arguments in double-quotes to make the shell see the variables as single arguments, in case they contain spaces.

answered Jan 30, 2013 at 10:15

Some programmer dude

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paxdiablo · Accepted Answer · 2013-01-30 10:15:57Z

5

Your arguments come from the command line rather than through standard input. Hence you would use:

./app "$file" "$text"

If it were coming in via standard input (one argument per line), you could use:

( echo "$file" ; echo "$text" ) | ./app

but that's not the case - you need to pass the arguments on the command line for them to show up in argv.

One other point, you'll notice I've put quotes around the arguments. That's a good idea to preserve whitespace just in case it's important. You should also do that in your lines:

path="$1"
ext="$2"
text="$3"

answered Jan 30, 2013 at 10:15

paxdiablo

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Comments

junix · Accepted Answer · 2013-01-30 10:15:51Z

2

Your invocation of the application is wrong. It should read

./app $file $text

answered Jan 30, 2013 at 10:15

junix

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