command line argument

Question

What is the output of this C code (run without any command line arguments)?
#include <stdio.h>

1.     int main(int argc, char *argv[])

2.     {

3.         while (*argv  !=  NULL)

4.         printf("%s\n", *(argv++));

5.         return 0;

6.     }

In this program argv gives the base address of array of char pointer so while(*argv !=NULL) condition is true and this should give the compile time error because of *(argv++) but when I execute this prog using cmd.... c:\ <prog_name>...output comes <prog_name> so how prog name is dispalying

Comments

Why do u think it will give compile time error? argv is array, and array can be qualified as pointer type as **argv

Answer 1

See here no error is present.

#include <stdio.h>
int main(int argc, char *argv[])
 {
    while (*argv  !=  NULL)// that means argv array is not reached end of string
    printf("%s\n", *(argv));// here pointer is pointing to one address
    argv++;// here it is pointing to next address
     return 0;
}

It contain program name because argv contains program name at element 0 and command line argument when argc is 1