Will the following give a compilation error?
delete cout;
delete cin;
The answer is : No.
It is a flaw in the implementation of stream classes from the Standard library. They have the following conversion function to void*
type, which means, all stream objects can be implicitly converted to void*
:
operator void * ( ) const;
This is very useful in general as it lets us write very idiomatic loop, say, when reading input from files. But at the same time, it lets user to write delete stream
. As I said, you can delete any stream object. So all of these are allowed:
delete ss; //declare std::stringstream ss;
delete iss; //declare std::istringstream iss;
delete oss; //declare std::ostringstream oss;
Only that they'll give a warning, saying (see at ideone):
warning: deleting ‘void*’ is undefined
which you can easily avoid just by casting, say, tochar*
. But the program has still issue, and most likely will crash when running it.
--
So my question is, has this issue been addressed and fixed, in C++11? The following article provides one fix for this problem:
--
Edit:
From @Xeo's comment on @Alf's answer:
- Is the safe-bool idiom obsolete in C++11?
The paper which proposed a fix for this issue:
- Explicit Conversion Operator
Draft Working Paper
Revision 1
It has apparently been fixed.
At least, in N3290 you have std::basic_ios::operator bool
instead of that void*
conversion, and this operator bool
is declared explicit
.
Note that C++98/C++03 did not support explicit
type conversion operators, but C++11 does.
An explicit
type conversion operator
N3290 §12.3.2/2;
is only considered as a user-defined conversion
for direct-initialization (8.5)
And that might seem to be impractical for the condition in e.g. a while
or for
statement.
Happily,
N3290 §4/3;
An expression e
can be implicitly converted to a type T
if and only if the declaration T t=e;
is well-formed, for some invented temporary variable t
(8.5). Certain language constructs require that an expression be converted to a Boolean value.
An expression e
appearing in such a context is said to be contextually converted to bool
and is well-formed if and only if the declaration bool t(e);
is well-formed, for some invented temporary variable t
(8.5). The effect of either implicit conversion is the same as performing the
declaration and initialization and then using the temporary variable as the result of the conversion.
where bool t(e);
is a direct-initialization.
E.g. you don’t have to explicit convert a stream object used as condition in a while
, because there is implicitly an explicit conversion (he he).
Unfortunately, searching N3290 I can’t find any list of the “certain language constructs” where this happens, but in comments to this answer JohannesD wrote:
Searched through the FDIS for “contextually”, and the whole list seems to be: if
, while
,
do
, for
, noexcept
, and static_assert
conditions; the first operand of ?:
; both
operands of &&
and ||
; and the operand of !
.
Cheers & hth.,
If I can give my 2 cents, I think the standard library "flawed" a bit, with all the good intentions.
The operator void*()
had been introduced to allow code like
while(stream)
or if(!stream)
or while(stream && ...)
, without giving an implicit access to integer arithmetic (that operator bool
whould have given).
In fact, this disable integer arithmetic, but gives access to pointer features (like delete ...).
Now, in C++0x, an explicit oeprator bool()
had been introduced. It doesn't implicitly give access to whatever feature, since it requires an implicit conversion. But ... wait a bit: 'while(bool(stream))' or even while(static_cast<bool>(stream))
are so wordy...
Operator ! is explicit, and
'while(!!stream)' looks so effective that I even wonder why not accept this as a paradigm:
If I want something to be explicitly converted into bool, I just provide an operator!()
and give to !
the memaning of "is not valid" and of !!
as "is valid".
Much safer then an implicit conversion and not uselessly wordy: after all !
exist from ever!