Attempting to pass Linux command output into a variable

Question

I have the following code:

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

int readCmd(char* cmd) {
    FILE *fp;
    char path[1024];

    /* Open the command for reading. */
    fp = popen(cmd, "r");
    if (fp == NULL) {
        printf("Failed to run command\n" );
        exit(1);
    }

    /* Read the output a line at a time - output it. */
    while (fgets(path, sizeof(path)-1, fp) != NULL) {
        printf("%s", path);
    }
    /* close */
    pclose(fp);
    return path;
}


int main( int argc, char *argv[] ) {
char* gp123;
char* gp124;


gp124=readCmd("cat /sys/class/gpio/gpio124/direction");
gp123=readCmd("cat /sys/class/gpio/gpio123/direction");
printf("123: %s \n124: %s \n",gp123,gp124);

}

With the following output:

out
in
123: in
124: in

As you can see, gpio pin 123 is set to 'in' and 124 is set to 'out'.

However, both gp123 and gp124 are being assigned 'in'. I am rusty with functions and C. Could you fellas help me out a little please?

More specifically, why is it failing to assign the returned value for each command I read? Is this a local/global variable issue?

Thanks!

EDIT:

#include <stdio.h>
#include <stdlib.h>
char* readCmd(char* cmd)
{
FILE *fp;
static char path[1024];

/* Open the command for reading. */
fp = popen(cmd, "r");
if (fp == NULL) {
    printf("Failed to run command\n" );
    exit(1);
}

/* Read the output a line at a time - output it. */
while (fgets(path, sizeof(path)-1, fp) != NULL) {
    printf("%s", path);
}
/* close */
pclose(fp);
return path;
}


int main( int argc, char *argv[] )
{
    char gp123[50];
    char gp124[50];
    char *answer;

    answer=readCmd("cat /sys/class/gpio/gpio124/direction");
    sprintf(gp124, "%s", answer);
    answer=readCmd("cat /sys/class/gpio/gpio123/direction");
    sprintf(gp123, "%s", answer);
    printf("123: %s \n124: %s \n",gp123,gp124);

}

This seems to work correctly, is there anything here I should change to be more correct?

try something like this: char gp123[50]; char gp124[50]; char *answer; answer=readCmd("cat /sys/class/gpio/gpio124/direction"); sprintf(gp124, "%s", answer); gp123=readCmd("cat /sys/class/gpio/gpio123/direction"); sprintf(gp123, "%s", answer); — Paulo
– Paulo, Commented Nov 13, 2015 at 18:11
I think the return of readCmd() is unique, so when it returns to gp123, the value of variable gp124 is changed to gp123's value. — Paulo
– Paulo, Commented Nov 13, 2015 at 18:12
Hey, Kizz, could you please, give me a like in my comment as it worked for you? =) — Paulo
– Paulo, Commented Nov 13, 2015 at 18:38
using the 'edited' version of the code: there is no way the compiler did not produce any warnings. The unused parameter: argc and the unused paramter: argv[] will both cause the compiler to produce a warning. When compiling, always enable all the warnings. (for gcc, at a minimum use: -Wall -Wextra -pedantic ) To fix the compiler warnings declare main() as: int main( void ) — user3629249
– user3629249, Commented Nov 14, 2015 at 16:05

Petr Skocik · Accepted Answer · 2015-11-13 18:12:46Z

1

You can't return a pointer to automatic memory from a function. The path array only exists for the duration of the readCmd function (which should return char*, not an int).

To solve it, you can:

declare the array static (or use a global array--pretty much the same thing) and embrace non-reenterability
use the heap (malloc the array or use the GNU getline function)
ask the caller to provide provide a pointer to where to save the output (its size should be passed in too)

answered Nov 13, 2015 at 18:12

Petr Skocik

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1 Comment

Kizz Over a year ago

I changed path to static and converted the function to a char*. This gave me the correct output and eliminated all compiler warnings.

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