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c++ templates - is this possible?

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Started by Nitage July 29, 2005 09:07 AM

13 comments, last by Nitage 18 years, 9 months ago

Nitage

1,107

Author

July 29, 2005 09:07 AM

say i have a class like this:


template<class num_t>
class quantity
{
public:
    template<class result_num_t, class other_num_t>
    quantity<result_num_t> operator*(const other_num_t& other)
    {
         return quantity<result_num_t>(_value * other.getValue());
    }
    num_t getValue(){return _value;};

private:
    num_t _value;
};

Is there any way to get the compiler to deduce the return type of operator* ? Now I believe I could write:



quantity<A> a;
quantity<B> b;

quantity<ABproduct> result = a.operator*<ABproduct>(b);

but the syntax is horrible and I have to explicitly state the return type, which isn't always convienient (such as in compound statements). Note: I've got boost installed and I'm prepared to delve as deeply into template 'magic' as is needed if this is possible. [Edited by - Nitage on July 29, 2005 9:42:04 AM]

SiCrane

11,840

July 29, 2005 09:11 AM

No, it's not possible. C++ will not do type deduction on the return type of a template function.

Edit: what you can do is return a type the relies on both types, like make a template type
template <typename A, typename B> ProductType
and return that or a type typedefed inside that type.

Nitage

1,107

Author

July 29, 2005 09:14 AM

Any way around this or recommendations on how I could improve the function?

At the minute I'm just using the tpye of the left hand argument.

As an aside, is there a proposal to change this in C++0x?

SiCrane

11,840

July 29, 2005 09:15 AM

AFAIK, there are no plans to change this in the next version of the C++ standard.

Nitage

1,107

Author

July 29, 2005 09:43 AM

Ok - slightly different question:

If I have two types, say A and B, is there any expression which gives me the return type of A*B?

mfawcett

374

July 29, 2005 01:37 PM

Quote:Original post by SiCrane
AFAIK, there are no plans to change this in the next version of the C++ standard.

I thought the 'auto' keyword was being thought about for this exact reason?

auto iter = myvec.begin();

The compiler has enough information, it just doesn't expose it through the language right now. This is one of the proposals that had a lot of support last time I checked. Has this changed? Also I haven't read it entirely, so I'm not sure if it includes templated functions or not, but as the types are replaced at compile time, I can't see how template functions provide any more of a problem that a normal function.

--Michael Fawcett

mfawcett

374

July 29, 2005 01:41 PM

Quote:Original post by Nitage
Ok - slightly different question:

If I have two types, say A and B, is there any expression which gives me the return type of A*B?

I'm using an older version of something that was in the boost sandbox for that, but I think it's been replaced by boost::mpl. I currently use something that looks like this:

boost::result_of_plus<A, B>::type
boost::result_of_multiplies<A, B>::type

and you can specialize the result_of structs for your own custom types as well. If you'd like the header I can send it to you, but it did not make it into any official boost release. Again, I think the functionality got folded into boost::mpl. Someone else may know for sure.

Here's an example (It's pretty ugly though).

template <typename LHS, typename RHS>inline typename boost::result_of_plus<LHS, RHS>::type	operator +(const LHS &lhs, const RHS &rhs){	typename boost::result_of_plus<LHS, RHS>::type nrv(lhs);	nrv += rhs;	return nrv;}

--Michael Fawcett

illone

151

July 29, 2005 01:51 PM

Quote:Original post by mfawcett
Quote:Original post by SiCrane
AFAIK, there are no plans to change this in the next version of the C++ standard.

I thought the 'auto' keyword was being thought about for this exact reason?

auto iter = myvec.begin();

the OP doesn't want the assigned-variable to be dependent on the function's return-type, he wants the function's return type to be dependent on the assigned-variable. more like:

auto my_function(int a)
{ ... }

Object b = my_function(3);

where my_function would somehow know to return a type 'Object'.

even if possible through some magic, i believe this would violate the more fundamental rule where a "over-loaded function cannot differ by only its return type."

illone

illone

mfawcett

374

July 29, 2005 02:21 PM

Quote:Original post by illone

the OP doesn't want the assigned-variable to be dependent on the function's return-type, he wants the function's return type to be dependent on the assigned-variable. more like:

Yes, of course. My misunderstanding.

--Michael Fawcett

iMalc

2,466

July 29, 2005 05:59 PM

When num_t, result_num_t, and other_num_t are supposed to be the same you can ditch result_num_t, and other_num_t and just use num_t.
It wouldn't even need to be a templated function, only a templated class.

But if you want to be able to multiply a 'quantity' by a 'foo_bar' then just write an explicit operator function for it and make foo_bar a friend, again without templates, making the return type obvious.
So basically you have to specify a * operator function for each type you can multiply 'quantity' by.

"In order to understand recursion, you must first understand recursion."
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c++ templates - is this possible?

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