CODEWITHSUNDEEP
cshellcommand
user976754
user976754

Reputation: 361

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

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

Upvotes: 1

Views: 2716

Answers (1)

Zombo
Zombo

Reputation: 1

"../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.

Upvotes: 2

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