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From: [e-mail] (Mike Rubenstein)
Newsgroups: comp.lang.c
Subject: Re: What is &Variable (declared as: char Variable[10])?
Date: Thu, 29 Feb 1996 20:46:43 GMT
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[e-mail] (Greg Goodrich) wrote:

> Mike Rubenstein ([e-mail]) wrote:
> : [e-mail] (Greg Goodrich) wrote:
>
> : > Abu Wawda ([e-mail]) wrote:
> : > : I'm having trouble understanding what the address of a static array
> : > : is. For example, if I declare a variable called myarray as:
> : > :     char myarray[10];
> : > : then what could &myarray possibly mean? myarray is not a pointer, so
> : > : &myarray could not possibly be the address of the variable myarray
> : > : (like it would be if I did char* myarray and then asked for &myarray).
> : >
> : > : Functions such as scanf() allow the following:
> : >
> : > :     char myarray[10];
> : >
> : > :     scanf("%s",&myarray);
> : >
> : > : but I don't understand what scanf() could possibly be taking in the
> : > : second parameter. It can't be: char** since myarray is not a
> : > : pointer. I CAN understand how the following would work:
> : >
> : > This is because C treats the occurrence of array names as the address of
> : > the array.  Therefore the following are equivalent:
> : >
> : >     scanf("%s", myarray);
> : >     scanf("%s", &myarray);
> : >
> : > The thing is, the address is implied when you use an array name.  This
> : > is not so for pointers.  I am not sure why C was incorporated in this
> : > way.  In my humble opinion, it would make it easier and clearer to force
> : > the programmer to put the & in front of array names, the same as you
> : > have to do for any other storage class.
>
> : No.  The are not equivalent.  The first is legal and the second is
> : not.
>
> : The %s format item in scanf expects a pointer to char.  myarray is
> : converted to a pointer to char so it is legal.  &myarray is a pointer
> : to array of 10 char and is not converted.  This results in undefined
> : behavior.
>
> : In many implementations pointer to char and pointer to array of 10
> : char have the same representation and this will work properly, but
> : this is not required by the standard.
>
> I would like to see an example of how this could be implemented to fail.
> They both point to the exact same memory address, the only difference
> being the datatype of the value, which can be type casted if necessary.
> If you have an array, the address of the array is also the address of
> the first member of the array, because the array itself is not a
> pointer, but a reference.  The same goes for a structure.  The address
> of a struct is the same as the address of the first member of the
> struct, but of different data type.  One is pointer to struct and the
> other is pointer to whatever the first member is declared as, but they
> are both pointers, and therefore are the same size.

Let me urge you to stop asking such questions.  The fact that the
standard says it is illegal is sufficient.  Guessing whether there is
an implementation in which it fails is fruitless since even if there
is not someone might write one tomorrow.

In any case, I find it hard to justify code based on lack of
imagination.

The construction is likely to fail in any implementation in which
sizeof(char*) != sizeof(char (*) 10) or in which they have different
representations.

Michael M Rubenstein