Passing a general argument to function using template

Question

There are some classes and their relationships are as follows:

class X:A,B
class Y:A,B
class Z:A,B

I want to pass a general type that will inherit from A and B to testFunction using template. My code is as follows:

template<template T:public A, public B>
void testFunction(T generalType)
{
    //do something
}

but my compiler told me that it's error template. How could I fix it?

Reopened this - the supposed duplicate was about a _class_. But unlike classes, functions have overloading. That means you want to _exclude_ specific functions from the overload set, rather then have them be selected in overload resolution and then break later on. — MSalters, Apr 26 '16 at 10:21
@ MSalters @ Aaron McDaid @songyuanyao. Thanks for all of your patient and careful answers. With all your help, my code pass the stage of compilation. I have learned a lot from all of you, — rontgen, Apr 27 '16 at 01:36

MSalters · Answer 1 · 2016-04-26T10:21:47.820

9

The standard method to conditionally define a template is std::enable_if<condition>. In this case, you want to check the condition std::is_base_of<A,T>::value && std::is_base_of<B,T>::value

edited Apr 26 '16 at 10:21

answered Apr 26 '16 at 10:19

MSalters

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score 4 · Accepted Answer · answered Apr 26 '16 at 10:23

template<template T:public A, public B> is not valid syntax. You could check the type with std::is_base_of and static_assert, which will trigger a compiler error if wrong type passed.

template <typename T>
void testFunction(T generalType)
{
    static_assert(std::is_base_of<A, T>::value && std::is_base_of<B, T>::value, "T must inherit from A and B.");
}

DEMO

Passing a general argument to function using template

2 Answers2