conversion operator - beginner's question

[subramanian100in]

In the book, C++ Coding Standards book by Hereb Sutter and Andrei
Alexandrescu, in Item 40 on pages 86-87 viz, "Avoid providing implicit
conversions", the authors have advised the use of named functions that
offer conversions instead of conversion operators.

In page 87, example 2: Errors that work.

class String
{
// ...
public:
operator cons char *(); // deplorable form
};

Assume s1, s2 are 'String's:
int x = s1 -s2; // compiles; undefined behaviour
const char *p = s1 - 5; // compiles; undefined behaviour
....

I am NOT going against the authors in asking the following.
However, as a beginner I just to want to know if conversion operators
should be avoided totally ?
Because, in <istream>, we have operator bool() const; (in the sentry
class)

This helps in
while(cin)
// ...

So here the standard library uses a conversion operator. Are there any
guidelines as to when the conversion operators can be defined ?

Kindly clarify.

I once again state that I am not confronting the authors. I am asking
this question from learner's point of view.

Thanks
V.Subramanian

[=?UTF-8?B?RXJpayBXaWtzdHLDtm0=?=]

On 2007-11-29 17:25, su************* *@yahoo.com, India wrote:

In the book, C++ Coding Standards book by Hereb Sutter and Andrei
Alexandrescu, in Item 40 on pages 86-87 viz, "Avoid providing implicit
conversions", the authors have advised the use of named functions that
offer conversions instead of conversion operators.

In page 87, example 2: Errors that work.

class String
{
// ...
public:
operator cons char *(); // deplorable form
};

Assume s1, s2 are 'String's:
int x = s1 -s2; // compiles; undefined behaviour
const char *p = s1 - 5; // compiles; undefined behaviour
...

I am NOT going against the authors in asking the following.
However, as a beginner I just to want to know if conversion operators
should be avoided totally ?

You should never say never, there are no absolute rules. Though in most
cases it is advisable to follow them, especially if you are a beginner.
But if you know what you are doing then there situations where they can
be useful.

Because, in <istream>, we have operator bool() const; (in the sentry
class)

This helps in
while(cin)
// ...

So here the standard library uses a conversion operator. Are there any
guidelines as to when the conversion operators can be defined ?

Kindly clarify.

One example where they might be useful is when creating proxy-objects,
then it might have a conversion to the object-type it is acting as a
proxy for.

--
Erik Wikström

[subramanian100in]

On Nov 29, 10:59 pm, "Alf P. Steinbach" <[e-mail] wrote:

* subramanian10.. [e-mail], India:

Because, in <istream>, we have operator bool() const; (in the sentry
class)

The sentry class is just a specification "implementation " detail, a
helper abstraction used to specify the functionality. You will never
use that operator bool() directly. Additionally, the modern templated
standard iostreams are not an example of good design. They're an
example of lack of design and/or bad design. Simply put, if some way of
doing things is employed by standard iostreams, then you know that it's
most probably something to not adopt in your own designs, that it's most
probably something to stay very very clear of, to fear and avoid.

This helps in
while(cin)
// ...

No, what's invoked here is "std::basic_ios <char>::operato r void*".

That's still an example of an Evil(TM) way of doing things.

With billions of good ways to do this, it might seem fantastic and
unbelievable that one of the very few really Evil ways ended up in the
standard, sort of like "we really need some more Evil, let's see...",
but such is standardization , and ask me not why: it's a mystery.

So here the standard library uses a conversion operator. Are there any
guidelines as to when the conversion operators can be defined ?

A conversion operator can implement a logical IsA relationship where
inheritance isn't applicable.

Cheers, & hth.,

- Alf

I found the following piece of code in C++ Primer 4th edition by
Stanley Lippman in page 26.

Sales_item trans;

while (std:cin >trans)
// ...

Here, isn't operator bool() called ?

Also let me know what is the correct way of doing the 'while
condition' if it is not a good way of doing.

Kindly clarify.

Thanks
V.Subramanian

[Kai-Uwe Bux]

su************* *@yahoo.com, India wrote:

On Nov 29, 10:59 pm, "Alf P. Steinbach" <[e-mail] wrote:

>* subramanian10.. [e-mail], India:

Because, in <istream>, we have operator bool() const; (in the sentry
class)

The sentry class is just a specification "implementation " detail, a
helper abstraction used to specify the functionality. You will never
use that operator bool() directly. Additionally, the modern templated
standard iostreams are not an example of good design. They're an
example of lack of design and/or bad design. Simply put, if some way of
doing things is employed by standard iostreams, then you know that it's
most probably something to not adopt in your own designs, that it's most
probably something to stay very very clear of, to fear and avoid.

This helps in
while(cin)
// ...

No, what's invoked here is "std::basic_ios <char>::operato r void*".

That's still an example of an Evil(TM) way of doing things.

With billions of good ways to do this, it might seem fantastic and
unbelievable that one of the very few really Evil ways ended up in the
standard, sort of like "we really need some more Evil, let's see...",
but such is standardization , and ask me not why: it's a mystery.

So here the standard library uses a conversion operator. Are there any
guidelines as to when the conversion operators can be defined ?

A conversion operator can implement a logical IsA relationship where
inheritance isn't applicable.

Cheers, & hth.,

- Alf

I found the following piece of code in C++ Primer 4th edition by
Stanley Lippman in page 26.

Sales_item trans;

while (std:cin >trans)
// ...

Here, isn't operator bool() called ?

Nope. As Alf said, the conversion involved goes to void*

Also let me know what is the correct way of doing the 'while
condition' if it is not a good way of doing.

Given the way iostreams handle things, this is a correct way to write the
while loop. However, that does not imply that it was a good idea to make
the standard streams support conversion to void*.
Best

Kai-Uwe Bux

[James Kanze]

On Nov 29, 5:25 pm, "subramanian10. [e-mail], India"
<subramanian10. [e-mail] te:

In the book, C++ Coding Standards book by Hereb Sutter and Andrei
Alexandrescu, in Item 40 on pages 86-87 viz, "Avoid providing implicit
conversions", the authors have advised the use of named functions that
offer conversions instead of conversion operators.

In page 87, example 2: Errors that work.

class String
{
// ...
public:
operator cons char *(); // deplorable form
};

Assume s1, s2 are 'String's:
int x = s1 -s2; // compiles; undefined behaviour
const char *p = s1 - 5; // compiles; undefined behaviour
...

I am NOT going against the authors in asking the following.
However, as a beginner I just to want to know if conversion operators
should be avoided totally ?

Pretty much so. About the only general exception I can think of
is proxies, which only work because of implcit conversions.

Otherwise, there are a few special cases; as a beginner, you
probably don't have to worry much about these, however.

Because, in <istream>, we have operator bool() const; (in the
sentry class)

This helps in
while(cin)
// ...

You're mixing up two things.

In istream and ostream, there is a nested sentry class, which
has an implicit conversion operator to bool. It is used when
writing << and >operators which go directly to the streambuf
(rather than decomposing the operation into << or >on simpler
types). The "standard" idiom here is:

std::ostream&
operator<<( std::ostream& dest, SomeType const& obj )
{
std::ostream::s entry s( dest ) ;
if ( s ) {
// do it...
}
return dest ;
}

Obviously, other alternatives (not involving the explicit
conversion) are possible, but in this (special) case, it really
doesn't matter; the class is designed to be used in one specific
case, and only in one specific case.

Both istream and ostream derive from ios, and ios has an
implicit conversion to void*. This supports idioms such as:
while ( std::cin >someVariable ) ...
or
while ( std::getline( std::cin, line ) ) ...
, the void* acting here as a bool. (The reason bool wasn't used
was because bool is an integral type, and >and << are defined
over it. So in some cases, you could get some strange overload
resolutions, rather than an error from the compiler.)

Whether this is good design or not is very debatable; as a
general rule, it is NOT a good idea to both modify program state
and have flow control in a single line/statement. Something
like:

std::cin >someVariable ;
while ( std::cin.succee ded() ) {
// ...
std::cin >someVariable ;
}

would arguably be better. Similarly, one would like to be able
to use the stream for initialization, and not just to modify
existing variables.

All one can say here is that I/O, in general, is hard, and that
iostreams works a lot better than anything else anyone has
proposed (or what's available in other languages). And the
idiom is ubiquious enough that it doesn't cause problems in
practice; everyone knows it and expects it. Since that's highly
unlikely to be the case for any class you write, you should
probably avoid such things.

So here the standard library uses a conversion operator.

The standard library does a lot of things that aren't
necessarily good practice. (The iostream's part is probably one
of the better parts of it.) But a standard library also obeys
different rules than your code, at least partially. It will be
ubiquious, and it will be well known. Which means that code
using it idiomatically will be readable, regardless of the
conventions. Your code won't have those advantages.

Are there any guidelines as to when the conversion operators
can be defined ?

Yes. Except for proxies and if you're designing the standard
library of a major language, don't. (As with all rules, there
are exceptions. But unless it is obvious that the case must be
an exception, it probably isn't.)

--
James Kanze (GABI Software) email:ja******* **@gmail.com
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