reading in arguments for execv()

[beatlea]

Hello,
I have to write a program using exec() where user can specify the pathname and arguments. I am trying to read in the arguments in the following way:

#include <stdio.h>
#include <unistd.h>
#include <sys/wait.h>

int main()
{
  ...
  int i;
  char *cmd[100];
  char argument[100];
  
  i = 0;
  for(i=0;i<3;i++)
  {
    printf("Enter %d. argument: ",i);
    scanf("%s",argument);
    cmd[i]=argument;
  }
  cmd[3] = (char *)0 ;

...

return 0;
}

But all I get is the last inputted argument in all 3 strings. I don't understand why. Could anyone help me with this?

Anna

[Aia]

char *cmd[100]; /* this is an array of pointers */
char argument[100]; /* this is just ONE array of 100 elements long */

cmd[i]=argument; /* Each time a new pointer element is pointing to the same array string */

You are changing the string every time through the loop. All those pointers points to the same string.

char argument[100] should be char argument[100][x]; /* where x is the length of the strings */

[beatlea]

I thought so, but if I do it as:

cmd[0]="some_string";
cmd[1]="another_string";
etc.

it works fine.

[Aia]

I thought so, but if I do it as:

cmd[0]="some_string";
cmd[1]="another_string";
etc.

it works fine.

Yes, why do you think it wouldn't work?
cmd[0] is pointing to a literal string in memory containing the "some_string"
cmd[1] is pointing to anoter literal string in memory containig the "another_string",
Two different places in memory. Now:
cmd[0] = argument; is pointing to argument[100] a variable string in memory.
cmd[1] = argument; is pointing to the same place in memory than the previous pointer is.
You change argument[100] and each pointer will still be pointing to whatever argument hold.

[Edit:]
Wait a minute. Are you trying to copy whatever is in argument[100] into cmd?

[beatlea]

I see. I used an array of strings as you adviced me and it works like this. Thank you.

Yes, I was trying to copy it. Is it fine or should I rather malloc cmd and point it to the input directly without the variables?

I mean I understand now that I'm not copying the string, but only its address in the memory. I just wonder if its better like this or without these variables.

[Aia]

>Yes, I was trying to copy it
Did you figure out why it would not work to try to copy into a pointer?
A pointer like cmd[0] can hold only the address location of some place in memory.
cmd[0] = argument; will hold only the address location of the first element of argument.

Things like cmd[0] = "Some_string" is deceiving in the way that it makes you think that
cmd[0] holds the value string. However all that is doing is pointing to the `S' of Some_string somewhere in `read only' memory. That's why you would never be able to change "Some_string".

>Is it fine or should I rather malloc cmd and point it to the input directly without the variables?
If you know how many strings you need I don't see why you need to go directly using malloc to do it for you.

[beatlea]

Yes, I understand it now. Thank you.

I don't know how many strings I need.

[Aia]

>I don't know how many strings I need.
Well then you might need to make use of malloc and create the room for it as you go into the heap memory.