pointers, arrays, char**, i don't know...

[landog]

I have been trying a number of ways to copy char *argv[].
I need to parse the command line and call different functions with a subset of the main argument list.

/* the objective is to copy argv into another array */
/* eventually, for example, copy the 3rd, 4th, 5th argv into 1st, 2nd 3rd arguments */

int main (int argc, char *argv[])
{
    int j;
    char * arguments[30];  /* max argument size */
    
    for (j = 1; j < argc; j++)
    {
        strcpy (arguments[j], argv[j]);  /* compiles, but get an unhandled exception here */
    }
}

int main (int argc, char *argv[])
{
    int j;
    char ** arguments;  
    
    for (j = 1; j < argc; j++)
    {
        strcpy (arguments[j], argv[j]);  /* Run-Time Check Failure #3 - The variable 'arguments' is being used without being initialized. */
    }
}

Any direction you can provide is much appreciated.

Thanks,
-dog

[Salem]

Try say

char arguments[30][100];

For your other attempts, you have to use malloc to allocate sufficient space.

[landog]

Try say

char arguments[30][100];

For your other attempts, you have to use malloc to allocate sufficient space.

Thanks. But, then how do I pass the arguments?

void my_function(int argc, char *argv[])
{
/*      ... etc., etc., etc.   */
}

int main (int argc, char *argv[])
{
    int arg_count = 0;
    char * arguments[30][30];  /* max argument size */
    
    for (int j = 1; j < argc; j++)
    {
        strcpy (arguments[j], argv[j]); 
        arg_count++'
    }

    my_function(arg_count, **arguments);   
/* argument of type "char" is incompatible with parameter of type "char **"  */
   
    my_function(arg_count,  *arguments);
/* argument of type "char *" is incompatible with parameter of type "char **" */

    my_function(arg_count,  arguments);
/* argument of type "char (*)[30]" is incompatible with parameter of type "char **" */
}

[rags_to_riches]

For one thing, compare the code Salem provided (ignoring the array length difference)

char arguments[30][100];

with yours

char * arguments[30][30];  /* max argument size */

You're also going to need another argument to your function.

void my_function(int argc, char *argv[], char my_arguments[][30])

[landog]

For one thing, compare the code Salem provided (ignoring the array length difference)

char arguments[30][100];

with yours

char * arguments[30][30];  /* max argument size */

Yeah, crud -- that was a typo. Should read:

char arguments[30][30];  /* max argument size */

You're also going to need another argument to your function.

void my_function(int argc, char *argv[], char my_arguments[][30])

There's not a way to call this?

void my_function(int argc, char *argv[])

Thanks for your help.
-dog

[claudiu]

You really need to read a tutorial about pointers and arrays in C before continuing this exercise. You seem to be confused about what a pointer is, what an array is and when is an array treated as a pointer to the first element.

I suggest you start here:

Cprogramming.com Tutorial: Pointers

and then here:

Cprogramming.com FAQ > Pointers And Arrays (intermediate)

As for you exercise if what you are trying to do is to call different functions depending on what the argument passed to the program is , then you probably need a something like a switch statement in main to branch away and call the appropriate function.

[tabstop]

If you do not need to alter the contents of the arguments, just which ones you have, then you just need to assign:

/* the objective is to copy argv into another array */
/* eventually, for example, copy the 3rd, 4th, 5th argv into 1st, 2nd 3rd arguments */

int main (int argc, char *argv[])
{
    int j;
    char * arguments[3];  /* I appear to want three arguments */
    
    for (j = 2; j < 5; j++)
    {
        arguments[j-2] = argv[j];
    }
}

If you intend to make a copy so that you can change things, you will need to acquire memory and then use strcpy to fill the memory.