Basic shell that takes multiple word command - c

Question

I am trying to implement a shell that takes a command ; however, I can not get it to work properly. For example, if I type "ls -a", I get this:
invalid option -- '
'
Try 'ls --help' for more information.
I have probably made some bad mistakes as I am a beginner so please forgive me. Also, I will put the code that reads in the command into a function. Its just like this for testing- thanks.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/wait.h>
#include <sys/types.h>
#include <unistd.h>
#include <stddef.h>
int main()
{
pid_t pid;
int status;
char* token;
char* argv[20];
char input[100];
printf("AP> ");
while (1)
{
fgets(input, 100, stdin);
token = strtok(input, " ");
int i = 0;
//walk through other tokens
while (token != NULL) {
argv[i] = malloc(strlen(token) + 1);
strncpy(argv[i], token, strlen(token));
//argv[i] = token;
i++;
token = strtok(NULL, " ");
}
argv[i] = NULL; //argv ends with NULL
pid = fork();
if (pid < 0)
{
perror("fork error");
return EXIT_FAILURE;
}
else if (pid == 0)
{
// child process
execvp(argv[0], argv);
perror("execl error");
return EXIT_FAILURE;
}
else {
// parent process
if (waitpid(pid, &status, 0)<0)
{
perror("waitpid error");
return EXIT_FAILURE;
}
}
printf("AP> ");
}
return 0;
}

Answer 1

If you insert the following loop after the inner while loop, then you'll see that you aren't removing the newline character at the end of input before the while loop starts.
for (i=0; argv[i]; i++)
printf("argv[%d] = '%s'\n", i, argv[i]);
Alternatively, you can use:
" \n"
as your separators instead of just
" "