Re: why does this work with Visual C++ 2005?
On Tuesday, 1 October 2013 01:53:26 UTC+1, Victor Bazarov wrote:
On 9/30/2013 7:45 PM, Lynn McGuire wrote:
Why does this work with Visual C++ 2005?
char buffer [10000];
void * test5 = & buffer;
I would think the compiler would give an error
on "& buffer", because "buffer" is the memory
address of the first element in the array, and
you can't get a memory address of a memory
address, there is no such thing.
The name of an array *decays* into a pointer to the first element of
that array in any expression *except* &<name>,
Or when it's the argument of sizeof, or when it's binding to
a reference parameter, or when it's an argument of typeid, or...
And of course, by "decays", you mean that there is an implicit
conversion which is used. Which in fact has nothing to do with
the fact that the expression is the name of an array, but only
with the fact that the type of the expression is an array type.
(The name of an array always has an array type, but there are
many other ways of obtaining an array type.)
where it's the address of
the entire array.
Just to be clear: the name of an array is never an address: it
has an array type. And when you take the address of an array
type, you get a pointer to an array (and *not* a pointer to the
first element, which you would get if the implicit conversion
took place).
Try wrapping a template around it to find out the
type of it.
Just doing "typeid(&buffer).name()" should be enough.
> Note that
test5 and buffer do point to the same memory
address in the VC++ debugger after execution of
that line of code.
Yes, the address of the array is the same as the address of its first
element, the only thing is different is the *type* of that pointer.
All data pointers convert implicitly to void*, and once it's
a void*, there's no trace of what the original type was.
Consider
char (*pb)[10000] = &buffer;
what's the type of 'pb'? Now, change the size either in the 'buffer'
declaration or here, do you get an error?
We do get a compiler error on:
char * test3 = & buffer;
And the following code compiles correctly:
char * test4 = & buffer [0];
The only thing i can think is the compiler
treats "& buffer" as "& buffer [0]" in the
void * case, assuming that is what the
programmer wanted.
Why would one think that? The only reasonable thing to think is
that &buffer is a pointer (the & operator always results in
a pointer type), but not a char*.
--
James