How to execute C inside Shell and capture the return value of C in Shell

jags1984 · 02-18-2014, 06:55 AM

Hi,

How to execute C inside Shell and capture the return value of C in Shell.

test.c

Code:

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

test.sh

Code:

./test hi how are you
echo $?

pan64 · 02-18-2014, 07:00 AM

c should be compiled and linked, the shell itself cannot do that.
You need to write that code into a file, compile it and execute from that script

allend · 02-18-2014, 07:11 AM

Use the system() function to set environment variable(s)?

rtmistler · 02-18-2014, 08:00 AM

What's your question? You wrote code which returns a value and you wrote your test.sh script to do exactly what you want. The fact that your code appears incorrect is part of this.

The function returns an integer, however argv[i] is a character array. Therefore you could either change it so that main returns char *, or you return i. By the way, i becomes the value of 5 therefore returning argv[5] is nothing.

Code:

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

Code:

~/testcode$ gcc -o test-return test-return.c
~/testcode$ ./test-return this is my function
argv[0] = './test-return'
argv[1] = 'this'
argv[2] = 'is'
argv[3] = 'my'
argv[4] = 'function'
~/testcode$ echo $?
5

I will grant that I do not know how to return a string from that function and have it be assigned to a bash variable.

metaschima · 02-18-2014, 10:48 AM

The program can only return an integer, which is used to signal if the program failed or not. That you can check with a bash if statement.

If you wanted to make the output of a C program into a bash variable:

Code:

var="$(./test hi how are you)"
echo "$var"