Obtaining the status of execv without stopping the execution of the parent - c

Question

Is there a way to obtain the status of a command run with execv without waiting for it with waitpid or wait?
The program I need to do reads other executable program as argument and need to print the status of this program when executed with execv exactly when the command is finished executing, but needs to keep executing after the fork instead of waiting for the program passed as argument to finish executing.
Here is what I have right now which doesn't seem to capture any status:
#define _GNU_SOURCE
#include <stdio.h>
#include <ctype.h>
#include <stdlib.h>
#include <unistd.h>
#include <stdio_ext.h>
#include <sys/wait.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <sys/shm.h>
int status;
void child_exit_handler(int signal)
{
int pid = wait(&status);
}
int main(int argc, char *argv[])
{
pid_t pid;
signal(SIGCHLD, child_exit_handler);
for (int i = 1; i < argc; i++)
{
pid = fork();
if (pid == 0)
{
char *cmd[4];
cmd[0] = "sh";
cmd[1] = "-c";
cmd[2] = argv[i];
cmd[3] = NULL;
execv("/bin/bash", cmd);
}
}
if (pid > 0)
{
// do something;
// later do something with status
}
}