Best alternative to a typedef for a function template?

Question

What I\'d like to do is something like this:

template 
DataType myFunc(DataType in)
{
   ...
}

typedef myFunc myFunc_i;

myFunc         


        

Answer 1

I would make a just tiny improvement over Xeo's answer.

Instead of using:

auto const myFunc_i = &myFunc<int>;

I would use

constexpr auto myFunc_i = &myFunc<int>;

Answer 2

Try this:

typedef int (*myFunc_i_type)(int);
myFunc_i_type const myFunc_i = &myFunc<int>;

This will trigger a compiler error if the signature of myFunc ever changes. Making it const also prohibits reassigning. Function pointers are also callable like any normal function: myFunc_i(5).

If you don't want a compiler error but rather have it update itself, use auto (again with const):

auto const myFunc_i = &myFunc<int>;

Answer 3

Some may disagree, some may jerk their knees against this, but consider this option:

#define myFunc_i myFunc<int>

As far as macros go, this one is quite safe; the only danger is that someone could redefine or undefine it later.

Aside from that, it has all the advantages of auto const myFunc = myFunc<int>;. If you want auto but don't want to use C++0x features yet, this is a more portable way to achieve the same effect.

Answer 4

Use decltype:

template <class DataType>
DataType myFunc(DataType in)
{
   ...
}

using myFunc_i = decltype(myFunc<int>(0));

myFunc_i(37);

Answer 5

Code Example:

#include <iostream>


template <typename T>
T sqr(T a){
    return a*a;
}

auto s = sqr<int>;
int main() {

    std::cout<<s(3.9);
    return 0;
}

prints 9, that means that used int function
auto requires c++0x, you may use real typename instead, but it's ugly and you'll need check changing signature too. Besides, this code, may disallow your compiler to inline this function

• As for me, it's often better use full name with <>

• In your case you need not use <> because myFunc is enough(DataType may be got from argument type)