Using pointers to iterate through argv[]

Given that the last element in argv is NULL, you don't need to index it or compare anything with argc if you really don't want to.

int main(int argc, char *argv[]) {
  for (char **arg = argv; *arg; ++arg) { // for each arg
    for (char *p = *arg; *p; ++p) {      // for each character
      process(*p);
    }
  }
}

*arg will be NULL at the end of the list of arguments, which is false. *p will be '\0' at the end of each string, which is false.

From N1256 5.1.2.2.1/2

If they are declared, the parameters to the main function shall obey the following constraints:

— The value of argc shall be nonnegative.

— argv[argc] shall be a null pointer.

Since for loop allows any kind of values, not necessarily integers for your "loop index", your loop could be rewritten like this:

for (char **a = argv ; a != argv+argc ; a++) {
    for(char *p = *a ; *p != '\0' ; p++) {
        // code uses *p instead of argv[i][j]
    }
}

The inner loop uses p as the loop variable, which is incremented with the regular p++, and checked with *p != '\0'. The loop condition could be shortened to *p, so the inner loop would look like this:

for(char *p = *a ; *p ; p++)

Yes you can iterate through argv using pointers.

The inner loop tells p to point at the beginning of argv+i and iterate through it until it reaches \0.

#include <stdio.h>
int main(int argc, char **argv) {
    int i;
    char *p;
    for(i=0; i < argc; i++) {
        for(p=*(argv+i); *p; p++)
            printf("%c", *p);
        printf("\n");
    }
}

If you are only interested in traversing the each arguments, but not interested in parsing each character, then you can simply.

#include <stdio.h>
int main(int argc, char **argv) {
    int i;
    char *p;
    for(i=0; i < argc; i++) {
        printf("Argument position %d is %s\n", i, *(argv+i));
    }
}