0

I want to know how an array of strings is declared? What I do is I declare an array of pointers of pointers to strings. Eg.

char *array[]= {"string1","string2","string3"};

I was reading about modifying environment variables in Linux and stumbled upon the pointer char **environ ( http://www.cs.bham.ac.uk/resources/courses/2005/17423/doc/libc/Environment-Access.html#Environment-Access ).

char **environ is declared as an array of strings. I think it should be a pointer to a pointer. For eg.

char *array[]= {"string1","string2","string3"};
 environ = array;

Am I doing something wrong?

I also read somewhere that char *argv[] = char **argv. How is it possible?

Edit: I also found this thread to be very helpful. Should I use char** argv or char* argv[] in C?

Community
  • 1
  • 1
Naruto Uzumaki
  • 958
  • 4
  • 17
  • 37
  • When I declared `char *array`, I was just giving an example as to what I believe is an array of pointers. I wanted to ask, what's exactly 'char **environ'? Is it a pointer to an array of pointer to strings or is it an array of strings? – Naruto Uzumaki Jan 17 '11 at 13:15

3 Answers3

1

well the problem is this. In your program are several pointer. One you asign to a array of strings and one called environ that points to the environment variables. What you say to C with environ = array is give environ the same value as array.. but array has a pointer to a local array. So after that statement the environ pointer will just point to the array you made but has not made any changes to its previous content.

I think you need to strcpy all elements of array to environ. Or use a api call setenv (i think it is)

and to you'r second question. Yes the first pair of [] can always be rewritten to a pointer. so array[] = *array as is array[][5] = (*array)[5] and there for *array[] = **array

i hope to have helped you.

Bram
  • 765
  • 1
  • 6
  • 14
  • When I declared `char *array`, I was just giving an example as to what I believe is an array of pointers. I wanted to ask, what's exactly 'char **environ'? Is it a pointer to an array of pointer to strings or is it an array of strings? – Naruto Uzumaki Jan 17 '11 at 13:14
1

you are mixing up two different things that are in fact difficult to know for someone who is learning C. Declaration of variables inside a function and as a function parameter are not the same thing. The equivalence 

char*argv[] ~~~ char **argv

holds because this a parameter (of main). There the array is in fact the same thing as declaring a pointer.

Your assignment environ = array is not wrong, syntactically, the compiler will accept it. But it is wrong semantically for several reasons:

  • You don't know who and how is allocated *environ.
  • You loose the reference to the initial contents of *eviron.
  • You assign a local storage that will be recycled once you leave the scope of the function. So *environ will be undefined, once you left the function.

So environ is a particularly bad example to do such an assignment.

Jens Gustedt
  • 76,821
  • 6
  • 102
  • 177
  • Aside from the array being in local storage, there's nothing wrong with assigning `environ` this way as long as you don't mix that with other functions for modifying the environment. It's probably the best and safest way to setup the environment for child processes, and as the first line of `main` in any suid-root program. – R.. GitHub STOP HELPING ICE Jan 17 '11 at 17:31
1

In C a string is basically just an array of chars. in addition an array name also represents its address.

this is the reason why argv[] is the address of the array of chars (which is a string) and *argv is also the address of the string (since it's the address of the first char).

SivGo
  • 226
  • 1
  • 2