argc and argv in main

Question

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAXLINE 1000

int getline(char *s, int lim)
{
    int i = 0;

    while(i < lim - 1 && (*s = getchar()) != EOF && *s++ != '\n')
       i++;

    if(*s == '\n')
      *s++ = '\n', i++;

    *s = '\0';
    return i;
}
int main(int argc, char *argv[])
{   

     char line[MAXLINE];
     long lineno = 0;
     int c, except = 0, number = 0, found = 0;

     while(--argc > 0 && (*++argv)[0] == '-')
        while(c = *++argv[0])
          switch(c) {
             case 'x':
                  except = 1;
                  break;
             case 'n':
                  number = 1;
                  break;
             default:
                  printf("find: illegal option %c\n", c);
                  argc = 0;
                  found = -1;
                  break;
          }

     if(argc != 1)
         printf("Usage: find -x -n pattern\n");
     else
         while(getline(line, MAXLINE) > 0) {
             lineno++;
             if((strstr(line, *argv) != NULL) != except) {
                   if(number)
                      printf("%ld:", lineno);
                   printf("%s", line);
                   found++;
             }
         }

     printf("Found: %d", found);
     return found;
}

while(--argc > 0 && (*++argv)[0] == '-')

Does this expression in the parantheses (*++argv)[0] differs from while(c = *++argv[0]) without parantheses?

If so, in what? Does (*++argv) means pointer to the next argument, and *++argv[0] means pointer to the next character in the current char array which is being pointed at?

Answer 1

yes, you are correct.

while(--argc > 0 && (*++argv)[0] == '-')

is scanning the array (of length argc) of command line arguments one by one looking for those starting with a - option prefix. For each of those:

while(c = *++argv[0])

is scanning through the set of switch characters that follow the first - in the current argument (i.e. t and n in -tn, until it hits the string null terminator \0, which terminates the while loop, since it evaluates as false.

This design allows both

myApp -t -n

and

myApp -tn

to both work and be understood as having the options t and n.

Answer 2

Incrementing argv is a very bad idea, as once you have done so it is difficult to get the original value back. It is simpler, clearer and better to use an integer index - after all argv IS an array!

To answer your question ++argv increments the pointer. This then has indirection applied to it to get the first character.

Answer 3

Yes, the two expressions differ (though only slightly). IMO, this code is a bit on the excessively clever side. You'd be better off with something like this:

for (int i=1; i<argc; i++)
    if (argv[i][0] == '-') {
       size_t len = strlen(argv[i]);
       for (int j=0; j<len; ++j)
           switch(argv[i][j]) {
               case 'x':
               // ...

This is pretty much equivalent to the code above, but I doubt anybody (who knows C at all) would have any difficulty figuring out what it really does.

Answer 4

The parentheses change the order in which the expressions are evaluated.

Without parentheses *++argv[0]:

1. argv[0] gets the pointer to character data currently pointed to by argv.
2. ++ increments that pointer to the next character in the character array.
3. * gets the character.

with parentheses (*++argv)[0]:

1. ++argv increments the argv pointer to point to the next argument.
2. * defereferences it to obtain a pointer to the character data.
3. [0] gets the first character in the character array.

Answer 5

I did find a 'bug' in this program tho, entering myprogram.exe -xxxxnnnnn in the command line will actually work(as intended lol?). Should i fix K&R's version about this?