hi,
If I declare a character array ( say, char arr[10] ), how do I see the
address of the first element of the array ? simply using cout << arr or
cout << &arr displays the entire string in the array, not the address.
thanks. 9 3739
Wondering Wanderer wrote: If I declare a character array ( say, char arr[10] ), how do I see the address of the first element of the array ? simply using cout << arr or cout << &arr displays the entire string in the array, not the address.
cout << static_cast<voi d*>(arr);
V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask
Wondering Wanderer wrote: If I declare a character array ( say, char arr[10] ), how do I see the address of the first element of the array ? simply using cout << arr or cout << &arr displays the entire string in the array, not the address.
That's because << overloads to operator<<(char *), which you don't need.
To get operator<<(unsi gned long), you must typecast:
cout << "0x" << hex << reinterpret_cas t<unsigned long>(arr);
I switched to hexadecimal to make the bit patterns in the pointer clearer.
Question for the crew; will this work?
typedef unsigned long ULong;
cout << "0x" << hex << ULong(arr);
--
Phlip http://www.greencheese.us/ZeekLand <-- NOT a blog!!!
Wondering Wanderer wrote: hi,
If I declare a character array ( say, char arr[10] ), how do I see the address of the first element of the array ? simply using cout << arr or cout << &arr displays the entire string in the array, not the address.
Try cout << static_cast<voi d*>(arr)
thanks victor, phlip and rolf. there's another typecasting way too :
cout << (int *) arr ;
Phlip wrote: Question for the crew; will this work?
typedef unsigned long ULong; cout << "0x" << hex << ULong(arr);
This conversion should require a reinterpret_cas t. VC++7.1 compiles it
with warnings though if one specifies compiler option /Wp64. This means
that the cast may change the value on a 64bit system.
typedef unsigned long long ULong;//...
....seems to make this warning go away.
see c++ standard98:
5.2.10/3 - mapping of reinterpret_cas t implementation defined
5.2.10/4 - A pointer can be explicitly (not implicitly) converted to
any integral type large enough to hold it.
Regards,
Werner -- Phlip http://www.greencheese.us/ZeekLand <-- NOT a blog!!!
werasm wrote: Phlip wrote:
Question for the crew; will this work?
typedef unsigned long ULong; cout << "0x" << hex << ULong(arr);
This conversion should require a reinterpret_cas t.
A C-style cast can do a reinterpret_cas t. By 5.2.3, that code is
equivalent to:
cout << "0x" << hex << (ULong)arr;
which is equivalent to:
cout << "0x" << hex << reinterpret_cas t<ULong>(arr);
Tom
Tom Widmer wrote: werasm wrote: Phlip wrote:
Question for the crew; will this work?
typedef unsigned long ULong; cout << "0x" << hex << ULong(arr);
This conversion should require a reinterpret_cas t.
A C-style cast can do a reinterpret_cas t. By 5.2.3, that code is equivalent to:
Yes, but that was not a c-style cast :-). That was instantiating a
temporary of type unsigned long...
Big difference between ULong(arr) and (ULong)arr, not true?
W
cout << "0x" << hex << (ULong)arr; which is equivalent to: cout << "0x" << hex << reinterpret_cas t<ULong>(arr);
Tom
werasm wrote: Tom Widmer wrote:
werasm wrote:
Phlip wrote: Question for the crew; will this work?
typedef unsigned long ULong; cout << "0x" << hex << ULong(arr);
This conversion should require a reinterpret_cas t.
A C-style cast can do a reinterpret_cas t. By 5.2.3, that code is equivalent to:
Yes, but that was not a c-style cast :-). That was instantiating a temporary of type unsigned long...
Big difference between ULong(arr) and (ULong)arr, not true?
No, the two are semantically identical - see 5.2.3 in the C++ standard.
For example:
typedef int* p;
int i = 1;
p pi = p(i); //meant to do p(&i)
Sadly, that is well-formed C++.
Tom
Tom Widmer wrote: No, the two are semantically identical - see 5.2.3 in the C++ standard. For example:
typedef int* p; int i = 1; p pi = p(i); //meant to do p(&i)
Sadly, that is well-formed C++.
Thanks (I've learnt), Yes truly sad (honestly - why?). In that case I'd
rather use:
cout << "0x" << hex << reinterpret_cas t<ULong>(arr); //...or better
std::cout << "0x" << std::hex << static_cast<voi d*>(arr) << std::endl;
Werner Tom This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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