How to pass shell variable as a command line arguments in C program

Question

I want to pass shell variable as command line arguments to C program. To practice this I have written a basic C program which is as follows:

#include<stdio.h>  
#include<stdlib.h>
int main(int argc, char **argv){  
    int i, a, prod=1 ;
    if(argc==2)
    a = atoi(argv[1]);
    else 
    a = 8 ; 
    printf("value of cmd: %d\n",a);
    for(i=1; i <= a ; i++)
        prod *= i ;
    printf("factorial %d!= %d\n",a, prod);
}

Now to pass the argument through shell script variable, I have written a small script which is as follows:

#!/bin/bash  
echo "Welcome to the small shell scripting!!!"
for ((i=4; i<9;i++))
do
        "../programming/ctest/arg $i"
done

Where ../programming/ctest/arg is the executable file for the above C program. But when I run the script output is ../programming/ctest/arg 4 no such file or directory and so on. while running without shell variable It gives the proper output. Kindly look at that, if this issue is solved I will use it for further automation

fuz · Accepted Answer · 2013-02-17 22:01:20Z

2

"../programming/ctest/arg $i"

You need to remove quotes good sir

../programming/ctest/arg $i

To elaborate, without removing the quotes Bash will interpret the entire string as a command, instead of a command and argument like you intended.

edited Feb 17, 2013 at 22:01

fuz

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answered Feb 17, 2013 at 20:02

Zombo

1

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2 Comments

user976754 Over a year ago

@DanielFischer, I could not get you. What do you mean by OP and it can't accept answer before the question is 15 minutes old? Kindly explain it. I just upvoted because the answer was useful for me.

user000001 Over a year ago

@user976754 It is even better if you quote the variable $i as in ../programming/ctest/arg "$i", in case the arguments contain spaces in them

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