type inference when using templates

问题

So here is what I would like to do: I use std::pair, but I would surely like to do the same using tuples, or indeed pretty much any kind of template. When assigning a pair variable, I need to type something like:

T1 t1;
T2 t2;
std::pair<T1,T2> X;
X = std::pair<T1,T2> (t1, t2);

Is there a way to omit the second <T1,T2> when creating the new pair, and let the compiler guess, either using X's type (I'm obviously trying to create a pair<T1,T2>) or t1 and t2's types (I am building a pair with a T1 object and a T2 object, there is a chance the pair I want is of type pair<T1,T2>) ?

回答1:

Yes, but template argument deduction only works on function templates, not constructors of class templates. For this reason, the library provides a function:

X = std::make_pair(t1, t2);

In C++11, pairs and tuples can be initialised and assigned from initialiser lists:

X = {t1, t2};

or auto can be used to automatically specify the type from the initialiser, so you don't need to specify the template arguments at all:

auto X = std::make_pair(t1, t2);

回答2:

Yes, use std::make_pair:

std::pair<T1,T2> X;
// some code
X = std::make_pair( t1, t2 );

Or, if you can initialize it directly:

std::pair<T1,T2> X = std::make_pair( t1, t2 ); // OR, even better:
std::pair<T1,T2> X( t1, t2 );

For C++11, initializing is even better:

auto X = std::make_pair( t1, t2 );

The assignment is the same:

X = std::make_pair( t1, t2 );

but it could also be:

X = { t1, t2 };

回答3:

std::pair<T1,T2> X = std::make_pair(t1, t2);

or with C++11

std::pair<T1, T2> X = {t1, t2};

or with C++11

auto X = std::make_pair(t1, t2);

And for std::tuple there is std::make_tuple.

来源：https://stackoverflow.com/questions/12192955/type-inference-when-using-templates

标签

c++

templates

types

inference