Question

#include <stdio.h>

 main(argc, argv)
 int argc;
 char *argv[];
 {
     register int i, nflg;

     nflg = 0;
     if(argc > 1 && argv[1][0] == '-' && argv[1][1] == 'n') {
      nflg++;
      argc--;
      argv++;     //Incements a constant pointer, how???
     }
     for(i=1; i<argc; i++) {
       fputs(argv[i], stdout);
       if (i < argc-1)
          putchar(' ');
     }
     if(nflg == 0)
       putchar('\n');
     exit(0);
 }

This program increments the value of argv, but argv is a constant pointer in C. Why don’t I get a compilation error from this?

Answer 1

First of all, the type of argv is char ** (remember that in the context of a function parameter declaration, T a[] is synonymous with T *a). Thus it’s a pointer type, not an array type, so use of the ++ operator is not immediately disallowed.

Secondly, while this looks like old-style K&R C, it’s still considered valid. Here’s what the C99 standard (n1256) says about argc and argv:

5.1.2.2.1 Program startup
…
2 If they are declared, the parameters to the main function shall obey the following
constraints:
…
— The parameters argc and argv and the strings pointed to by the argv array shall
be modifiable by the program, and retain their last-stored values between program
startup and program termination.

So the expression argv++ is perfectly legal.