question regarding char *argv[]

[cnb]

Hi,
I have question regarding following example from cprogramming faq section:

#include <stdio.h> 

int main(int argc, char *argv[])
{
  int i, j;
  for (i = 0; i < argc; i++)
  {
    printf ("argv[&#37;d] is %s\n", i, argv[i]);
      printf ("argv[%d][%d] is %c\n", i, j, argv[i][j]);
  }
  return(0);
}

/*
 * When invoked with:
 E:\>a.exe -parm1 -a
 * The output is:
  argv[0] is E:\a.exe
  argv[0][0] is E
  argv[0][1] is :
  argv[0][2] is \
  argv[0][3] is a
  argv[0][4] is .
  argv[0][5] is e
  argv[0][6] is x
  argv[0][7] is e
  argv[1] is -parm1
  argv[1][0] is -
  argv[1][1] is p
  argv[1][2] is a
  argv[1][3] is r
  argv[1][4] is m
  argv[1][5] is 1
  argv[2] is -a
  argv[2][0] is -
  argv[2][1] is a
 *
 */

I have following questions:
My understanding is that char *argv[] is a array of pointers to string. Is this a two dimensional array? So the *argv[] array contains pointers to string so thats one set of array but i cannot understand how is the second array works?
Is there a diagram that illustrates the char *argv[]?

Need to understand the following lines:
for (j = 0; argv[i][j]; j++)
printf ("argv[%d][%d] is %c\n", i, j, argv[i][j]);
In this case the code loops through for a value of i and keeps incrementing value of j. My question is how does it break out of this loop i.e. j=0 and than you increment to j=1 and j=2.....how does it stop this loop? Does it stop after it hits a null or something?

Thanks

[C_ntua]

argv[] is an array that stores all the arguments. The total number of the arguments are argc. So argv has argc rows. Now, the argument is a string. A string is a char*. Nothing else.
It is a 2D array, since a pointer is an array in a way. But one dimension is equal to argc, the other doesn't have a specific size, since each argument will differ in size. Some may right it like: int main(int argc, char** argv); which is the same. Generally [] means * in most cases in C.
The loop seems wrong for me! Are you sure you are not forgetting an inner loop?
In any case this:

  for (i = 0; i < argc; i++)
    printf ("argv[&#37;d] is %s\n", i, argv[i]);

Will print all the arguments. This:

  for (i = 0; i < argc; i++)
      for (j=0; argv[i][j] != \'0'; ++j)
          printf ("argv[%d][%d] is %c\n", i, j, argv[i][j]);

Would do the same, printing one by one character

Diagram for char* argv[]. Lets say you put on the command line >>P.exe 12

[P] [1] [2]
[. ] [\0] [\0]
[e]
[x]
[e]
[\0]

Now, argc will be 2. argv will be the above

[laserlight]

Diagram for char* argv[]. Lets say you put on the command line >>P.exe 12

[P] [1] [2]
[.] [\0] [\0]
[e]
[x]
[e]
[\0]

Now, argc will be 2. argv will be the above

That looks like a typo, plus argv actually has a terminating null pointer. More accurately:
[P] [1] [NULL]
[.] [2]
[e] [\0]
[x]
[e]
[\0]

Despite the three elements of argv, argc is indeed 2 in this case.

Whether argv[0] is "P.exe" or some other representation of the program name is implementation defined (and argv[0] can simply be left as an empty string).

[cnb]

Hi,
Yes there is typo in my original message:
The code looks like this:

#include <stdio.h> 

int main(int argc, char *argv[])
{
  int i, j;
  for (i = 0; i < argc; i++)
  {
    printf ("argv[%d] is %s\n", i, argv[i]);
    for (j = 0; argv[i][j]; j++)
      printf ("argv[%d][%d] is %c\n", i, j, argv[i][j]);
  }
  return(0);
}

/*
 * When invoked with:
 E:\>a.exe -parm1 -a
 * The output is:
  argv[0] is E:\a.exe
  argv[0][0] is E
  argv[0][1] is :
  argv[0][2] is \
  argv[0][3] is a
  argv[0][4] is .
  argv[0][5] is e
  argv[0][6] is x
  argv[0][7] is e
  argv[1] is -parm1
  argv[1][0] is -
  argv[1][1] is p
  argv[1][2] is a
  argv[1][3] is r
  argv[1][4] is m
  argv[1][5] is 1
  argv[2] is -a
  argv[2][0] is -
  argv[2][1] is a
 *
 */

Yes there is another for loop, but i still do not how this is working i.e. the second for loop:
The for loop i am talking about is:

for (j = 0; argv[i][j]; j++)

How does the expression arg[i][j] produces a false which means that you break out of this loop...i.e. what is the false condition.
This code is provide at the following URL:
http://faq.cprogramming.com/cgi-bin/...&id=1043284392
Its the last example on this page that i am looking at.
Thanks

[Dino]

Since strings (null terminated char arrays) are passed, when j hits the \0 (the null), it evaluates to false.

[robwhit]

'\0' evaluates to false. argv[i][j] is a nul terminator ('\0') when j is the last element of the array argv[i].

[Elysia]

Basically, it turns the expression into a boolean expression.
All expressions can basically be dumbed down to non-0 and 0.
0 means false, non-0 means true.
So when argv[i][i] is '\0', it evaluates to 0 which evaluates to false and breaks the loop.