array question

the name of the array is the first item address

Paul Rice,
your check is incorrect. using a char* type with cout displays
the string itself. check the following and see the same addresses:

char array[] = {'a','b','c','d','e','f','g','h','i','\0'};
cout << "string:\t" << (void*) array << endl;
cout << "address:" << (&array) << endl;

Paul Rice,
Sorry too about the confusion.
bottom line is array and &array are the address of the first element.

Regards,
Guy

'array' is really a constant (address to first element). You can't "take" an address of a constant, thus &array is wrong!

Originally posted by j0nas
'array' is really a constant (address to first element). You can't "take" an address of a constant, thus &array is wrong!

No, array is really the whole array. If 'array' were just an address, then 'sizeof(array)' would evaluate to the size of an address (or the size of a pointer). Instead, 'sizeof(array)' evaluates to the size of the entire memory of the array.

And why is '&array' wrong? I believe this is acceptable C (and C++) syntax to take the address of an array (I wish I had a copy of K&R around). It just returns the address of the first element (which makes sense if you think about memory layout). The fact that you can take the address of an array makes writing generic macros and templates much easier.

- Kevin

I do not have a copy of K&R around, but I'm holding
"A Book On C(Al Kelley,Ira Pohl) / Addison Wesley",
page 254-5:

say we have -
int a[N],*p;

"...Note that because a is a constant pointer, expressions such as
a=p
++a
a+=2
&a
are illegal. We cannot change the value of a."

illegal or not, I checked with .c extention and it is compiled.
back to array and &array, they are different types. the following
overload demonstrate it:

#include <iostream>

void test1(char (*)[10]);
void test2(char *);

int main()
{
char str[10]="12345";
test1(&str);
test2(str);

return 0;
}

void test1(char (*p)[10])
{
cout << "in test1" << endl;
cout << "*p=" << *p << " sizeof(*p)=" << sizeof(*p) << endl;

}

void test2(char* p)
{
cout << "in test2" << endl;
cout << "*p=" << *p << " sizeof(*p)=" << sizeof(*p) << endl;
}

Isn't it amazing how the simplest things can generate such interest.
:)

Anyways, I would see int array[10]; as a C++ object.
The type of this object is int[10], and int[10] has an operator int* built into it.
That is to say, it will implicitly convert from an int[10] to an int*.
Also, &array is a pointer to an int[10] object, and this pointer will also implicitly cast to an int*.

Ooops.

You are absolutely correct:
VC2003:
error C2440: 'initializing' :
cannot convert from 'int (*)[10]' to 'int *'
Types pointed to are unrelated; conversion requires
reinterpret_cast, C-style cast or function-style cast

Curious.