ofstream is_open

[Steven C]

ofstream blah;

IS:

if (blah == NULL)
{
}

THE SAME AS:

if (!blah.is_open())
{
}

I think the answer is NO since blah will never equal NULL, since I
could not find a operator for ofstream for == and blah is a instance
of a class and not a reference or a pointer therefore the statement
blah == NULL makes no sense.

 

[Ron Natalie]

I think the answer is NO since blah will never equal NULL, since I
could not find a operator for ofstream for == and blah is a instance
of a class and not a reference or a pointer therefore the statement
blah == NULL makes no sense.

The base class for the streams overload both a conversion operator for void*
and also operator!. Hence you can say things like:
if(blah)
or
if(!blah).

These are mapped to the blah.fail() (which emcompasses more errors than not
being
open).

 

[Sam Holden]

ofstream blah;

IS:

if (blah == NULL)
{
}

THE SAME AS:

if (!blah.is_open())
{
}

I think the answer is NO since blah will never equal NULL, since I
could not find a operator for ofstream for == and blah is a instance
of a class and not a reference or a pointer therefore the statement
blah == NULL makes no sense.

ofstream has a conversion operator to void*, so comparing it with NULL
is perfectly valid.

However, they have completely different meanings, the first checks if
failbit or badbit are true, the second tests if blah.is_open() returns
false. Those are completely different tests.

Hence the answer is no, but for a different reason.