c programming using execvp

Question

I'm working on a side project where I want to spawn 5 children processes then execvp into a second program called isEven which will tell if the number passed to it is even or not. However, the execvp call keeps failing as I ran it in gdb and it doesn't execute. Args and load(args) are both called in another function.

Thanks in advance!!!

//Code to load binary
void load(char* args[])
{
    execvp("./isEven", args);
    printf("ERROR\n");
} 

//args being passed in
char *args[] = {"./isEven", number, NULL};
load(args);

Here is a smaller program I wrote which does the same thing I'm trying to do right now. Gdb says I am getting a segmentation fault.

int main(int argc, char *argv[])
{
    int returnCode = fork();

    if(returnCode == 0)
    {
        char *args[] = {"./isEven", 12345, NULL};
        execvp("./isEven", args);
        printf(errno);
        printf("ERROR: execvp failed\n");
        exit(-1);
    }
    else if(returnCode < 0)
    {
        fprintf(stderr, "fork failed\n");
    }
    else
    {
        //i am the parent
        printf("I am the parent");
    }
    return 1;
}

Answer 1

The main problem is with this line:

char *args[] = {"./isEven", 12345, NULL};

The args passed to execvp must be an array of char*. Having 12345 as an argument makes it interpret it as address and it's unlikely to be a valid address. In any case, that's not what you wanted to pass. So do:

char *args[] = {"./isEven", "12345", NULL};

You need to convert number into a string. For example:

int number = 12345;
char num_str[256];
snprintf(num_str, sizeof num_str, "%d", number);
char *args[] = {"./isEven", num_str, NULL};
execvp("./isEven", args);
perror("exec");

---

In your modified example, you are printing errno as:

printf(errno);

This will fail because printf expects a string as its first argument. You can use perror() to print the error as shown above.