argv weirdness

Hi,

I'm experimenting with argv to learn more about how it works for a program I'm writing. I'm working with the code below:

Code:

---

#include<stdio.h>

int main( int argc, char **argv)
{
    char *array = argv[1];
    int i;
    printf( "%d\n", sizeof(array));
    for( i = 0; i < sizeof(array); i++ )
    {
        printf( "%c\n", array[i] );
    }
    return(0);
}

---

The thing that has me is the following behaviour:

Quote:

---

[calef13@bluenode c]$ ./test1 127.0.0.1 1-20
4
1
2
7
.
[calef13@bluenode c]$

---

Can anybody here explain why it's stopping at the first 0? I can't for the life of me fathom it. But I suspect I'm doing something undefined. The code above works just fine for a second parameter "1-200". Also if there is a better way to get a string out of argv so it can be compared character by character, let me know.

Put the IP address in double quotes.

EDIT - argv is an array of strings, not characters.

Quote:

---

Originally Posted by Calef13

Hi,

I'm experimenting with argv to learn more about how it works for a program I'm writing. I'm working with the code below:

Code:

---

#include<stdio.h>

int main( int argc, char **argv)
{
    char *array = argv[1];
    int i;
    printf( "%d\n", sizeof(array));
    for( i = 0; i < sizeof(array); i++ )
    {
        printf( "%c\n", array[i] );
    }
    return(0);
}

---

---

Also, your loop construct is wrong. sizeof(array) will most likely always be 4, since a pointer is most always 4 bytes. What you want is to loop based on argc - 1.

sizeof(array) is not a good test for a terminal condition as Dino pointed out.
walk along the char array that argv[1] points to until a NULL (string terminator) is seen.

ok thanks for all the replies everyone, I understand now, I should have been using strlen() for one. Just for completeness, the code is below.

Code:

---

#include<stdio.h>
#include<string.h>

int main( int argc, char **argv)
{
    char *array = argv[1];
    int i = 0;
    printf( "&#37;s\n", array );
    printf( "%d\n", strlen(array) );
    while( array[i] != '\0' )
    {
      printf( "%c", array[i]);
      i++;
    }
    getch();
    return 0;
}

---

So argv[1] is just a pointer to c string, nothing special. Python has ruined me :P

Note that argv[argc] is a null pointer, hence if argc == 1, argv[1] is a null pointer.

Quote:

---

Originally Posted by Calef13

ok thanks for all the replies everyone, I understand now, I should have been using strlen() for one. Just for completeness, the code is below.

---

Neither do you need strlen() nor do you need the variables i and *array (unless you don't want to change what argv[1] is currently pointing to).
Just increment argv[1] while printing out each character it points to in turn until you reach the end of the string signaled by the NULL character.

Quote:

---

Originally Posted by Calef13

Code:

---

#include<stdio.h>
#include<string.h>

int main( int argc, char **argv)
{
    char *array = argv[1];
    int i = 0;
    printf( "%s\n", array );
    printf( "%d\n", strlen(array) );
    while( array[i] != '\0' )
    {
      printf( "%c", array[i]);
      i++;
    }
    getch();
    return 0;
}

---

So argv[1] is just a pointer to c string, nothing special. Python has ruined me :P

---

yes argv[1] is just a pointer to an array of char.

You can use strlen to find the length of a string. You can print a string with %s.