1

So for example I want to create a function that add two numbers and return total.

template<typename T1, typename T2>
T1 add( T1 n1, T2 n2 ){
    return n1 + n2;
}

Problems is if T1 is int and T2 is float. Then function will return int. But I want it to return float. Is there a trick or a way to achieve it?

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4 Answers 4

Reset to default
10

If you are using C++11

decltype is your friend

template<typename T1, typename T2>
auto add( T1 n1, T2 n2 ) -> decltype(n1 + n2) {
    return n1 + n2;
}

This will use the type resulting from n1 + n2

More at http://en.wikipedia.org/wiki/C%2B%2B11#Alternative_function_syntax

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4

Yes

template<typename RT, typename T1, typename T2>
RT add( T1 n1, T2 n2 ){
    return n1 + n2;
}

Now call it like:-

add<float>(2,3.0);
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1 Comment

What would be syntax for arithmetic operators?
3

If you are using C++14:

template<typename T1, typename T2>
auto add(T1 n1, T2 n2)
{
   return n1 + n2;
}

This is like the C++11 version, but does not require you to write out the trailing-return-type yourself (which is nice).

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2 Comments

Why not decltype(auto)? add should return a reference if operator+ for two types returns one
@Columbo: I confess I know relatively little about this stuff. Please feel free to suggest an appropriate remedy.
1

Another C++11 solution would be to use std::common_type<T1, T2>::type as return type.

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