The prototype of execvp is
int execvp(const char *file, char *const argv[]);
It expects a pointer to char as the first argument, and a NULL-terminated pointer to an array of char*. You are passing completely wrong arguments.
You are passing a single char as first argument and a char* as the second.
Use execlp instead:
int execlp(const char *file, const char *arg, ...
/* (char *) NULL */);
So
char *token = strtok(input," \n");
if(token == NULL)
{
fprintf(stderr, "only delimiters in line\n");
exit(1);
}
if(execlp(token, token, NULL) < 0){
fprintf(stderr, "Error in execution: %s\n", strerror(errno));
exit(1);
}
Also the convention in UNIX is to print error messages to stderr and a process with an error should have an exit status other than 0.