execvp in C not going through ar - c

Question

I'm trying to use exec to execute a list of commands given as arguments.
Example input when In run the program would be ./assn2 ls date.
When I do this only the first command is executed.
#include<stdio.h>
#include<stdlib.h>
#include<sys/wait.h>
#include<unistd.h>
int main(int argc, char *argv[])
{
int args = argc-1;
pid_t childpid = fork();
// error
if (childpid < 0)
{
perror("fork() error");
exit(-1);
}
// parent process
if (childpid != 0)
{
printf("Parent Process started, now waiting for ID: %d\n", childpid);
wait(NULL);
printf("Parent Process resumeed. Child exit code 0. Now terminating\n");
exit(0);
}
// child process
if (args > 0)
{
printf("Child process has begun. %d argument/s provided\n", args);
int i;
for (i = 1; i <= argc; i++)
{
execlp(argv[i], argv[i], NULL);
}
execvp(argv[1], argv);
}
else
{
printf("No arguments provided, terminating child\n");
}
return 0;
}

Answer 1

Once the first child process execs (and succeeds), the for loop no longer continues because the an execlp would just replace the current process image with the command being exec'ed.
What you want to do is to loop over the command line arguments in the parent process and exec once for each of the command. Something like is probably what you're after:
for(int i = 1; i < argc; i++) {
pid_t pid = fork();
if (pid == 0) {
execlp(argv[i] ,argv[i], (char*)0);
perror("exec");
} else if (pid > 0) {
wait(NULL);
} else {
perror("fork");
exit(1);
}
}

Answer 2

What are you trying to achieve with the sequential calls to execlp() and execvp()? These functions are not meant to return. I think you should read the ref:
The exec() family of functions replaces the current process image with a new process image. [..] The exec() functions only return if an error has occurred.
As a result you cannot execute them one after another in the same process.
Read about fork():
fork() creates a new process by duplicating the calling process.
Moreover, here:
for(i = 1; i <= argc; i++)
you go out of bounds, since argv starts indexing from 0, and ends at argc - 1.
Chnage it to:
for(i = 1; i < argc; i++)