more argv and argc

Hi all,
I am just trying to write a program that will echo the command line minus the program name... here is my code, any idea why it doesn't work? Thanks!

Code:

---

#include <stdio.h>

int main(int argc, char * argv [])
{
  argv = argv+1;
  printf(" \n");

while(**argv != NULL)
  {
    while (**argv != NULL)
    {
      putchar(**argv);
      ++(*argv);
    }

        argv = argv + 1;

  }

        return 0;
}

---

Good question. The first argument in main (int argc) is a counter of the number of different statements that are in the command line. Here's an example of what i'm trying to say:
c:\>my_program open whatever.txt
In this example argc will be 2 (3 arguments starting at 0). Looking for char **argv to be null can have errors when ran. So instead count argc down to 1 (0 is the path).

Hi,
Why you are complicating things try this one.


By
A.Sebasti...


#include<stdio.h>
int main(int argc,char * argv[])
{
int i=1;
printf("<%d>\n",argc);
while(argc!=1)
{
printf("<%s>\n",argv[i]);
i++;
argc--;
}
}

I didn't even think about using argc! Thanks for the help.

Clarinetster

Code:

---

int main(int argc, char * argv [])
{
        int i;
        for(i=1;i<argc;i++) printf("%s",argv[i]);
        return 0;
}

---

Code:

---

#include <stdio.h>

int main (int argc, char * argv [])
{
 printf ("\n");
 argv++;
 while (*argv != NULL)
 {
  //printf ("%s ", *argv);
  //argv++;
  // reformatted so it looks like yours...
  while (**argv != "")
  {
  putchar (**argv);
  (*argv)++;
  }
  argv++;
 }
 return 0;
}

---

The problem with your first code wat that is tested whether **argv was NULL, which **argv is a char, so it doesn't make sense.

We had a question at UNI for this ages and ages ago. We also had to print them in reverse order and test if they used it wrong. I did the part your after like Witch_King. It is easier to understand and looks neater