Using execv (C language) to run commands from a linux command prompt

Question

The only part I am confused on thus far is how to set up execv with the first parameter as the current working directory. I've tried both "." and "~", neither are executing anything to the screen; same for "/." and "/~". I'm confused on how to have execv run something like this:

$ ./prog ls -t -al

And have it execute the commands after the program execution (which are stored into argv) in the current directory, or the same directory as the file is in (which will vary based on who is using it.)

My code:

#include <stdlib.h>
#include <stdio.h>
#include <unistd.h>
#include <sys/types.h>

void main(int argc, char *argv[])
{
    int pid;
    int count = 0;
    char *argv2[argc+1];

    for(count = 0; count < argc-1; count++){
        argv2[count] = argv[count+1];
        printf("Argv2: %s\n", argv2[count]);  //just double checking
        argv2[argc-1] = NULL;
    }

    pid = fork();
    if(pid == 0){
        printf("Child's PID is %d. Parent's PID is %d\n", (int)getpid, (int)getppid());
        execv(".", argv2);       //<---- confused here
    }
    else{
        wait(pid);
        exit(0);
    }
}

Some sample output:

$ ./prog ls -t -al
Argv2: ls
Argv2: -t
Argv2: -al
Child's PID is 19194. Parent's PID is 19193

Answer 1

I guess execv is what is required to be used. execvp is a lot nicer since it will look for commands in your PATH setting.

execv(".", argv2);       //<---- confused here

...

#include <errno.h>
#include <string.h>
if ( execv(argv2[0],argv2) )
{
    printf("execv failed with error %d %s\n",errno,strerror(errno));
    return 254;  
}

---

wait(pid);

...

pid_t wait_status = wait(&pid);