Template question: list<Bar<double>> ...

[janzon]

Hi

I have a class Foo that I use both as a Foo<double> and a Foo<float>. I
want to put instances of this class in a list container. The list
container itself is a template and need a type argument. The naive way
below does not work.

list<Bar<double>> myList;

It produces the error message: Error: "," expected instead of ">>". How
do I solve this problem?


Thanks in advance, Daniel

 

[Thomas Tutone]

Hi

I have a class Foo that I use both as a Foo<double> and a Foo<float>.

Those are two _different_ classes, instantiated from the same class
template.

I want to put instances of this class in a list container.

Since they are two different classes, you can't put both in a single
list container without resorting to some tricks that probably would
make this more complicated than it ought to be.

The list
container itself is a template and need a type argument. The naive way
below does not work.

list<Bar<double>> myList;

Change the above line to:

It produces the error message: Error: "," expected instead of ">>". How
do I solve this problem?

Well, that will solve _one_ of your problems.

Best regards,

Tom

 

[Michael]

It produces the error message: Error: "," expected instead of ">>". How
do I solve this problem?

The lexer is reading >> as one token. You need to put a space between
them:
list<Bar<double> > myList;

Michael

 

[Thomas Tutone]

Change the above line to:

std::list<Bardouble> > myList; // note space between the two ">"

Sorry, typo. Should be:

std::list<Bar<double> > myList; // note space between the two ">"

Best regards,

Tom

 

[janzon]

Change the above line to:

std::list<Bardouble> > myList; // note space between the two ">"

Adding the missing '<' it works perfectly. So one need to add a space
not to confuse the compiler? I mean, so it doesn't think you want to
use the >> stream operator.

Well, that will solve _one_ of your problems.

Well, the other is quite similar

 

[Marcus Kwok]

Adding the missing '<' it works perfectly. So one need to add a space
not to confuse the compiler? I mean, so it doesn't think you want to
use the >> stream operator.

As of now, yes. Usually, when it requires it, I will also add an extra
space at the beginning for symmetry:

std::list< Bar<double> > myList;

IIRC, this will be fixed in C++0x.

 

[Pete Becker]

I have a class Foo that I use both as a Foo<double> and a Foo<float>.

You have a template named Foo, from which you create two classes,

I
want to put instances of this class in a list container. The list
container itself is a template and need a type argument. The naive way
below does not work.

list<Bar<double>> myList;

It produces the error message: Error: "," expected instead of ">>". How
do I solve this problem?

typedef Foo<double> descriptive_name;
list<descriptive_name> myList;

or:

list< Foo<double> > myList;

or wait for a C++0x compiler, which will be available sometime in the
next ten years, and will accept:

list<Foo<double>> myList;

--

-- Pete

Author of "The Standard C++ Library Extensions: a Tutorial and
Reference." For more information about this book, see
www.petebecker.com/tr1book.