How to use template when number of arguments are not known?

Question

I have refered this link to meet my requirement . But when I am trying to implement the same in my code, it is throwing an error.

template<typename T, typename... Args>
void fun(T t, Args... args)
{
    cout << t;
}
int main()
{
    fun(1, 2.0, "Ranjan", "hi");//Error happens here
return 0;
}

The error at fun() is template<class T, <error type>>

What is going wrong here?

compiles for me fine (with some warnings), and executes good. What is the error? — BЈовић, May 15 '13 at 11:19
Compiles fine with g++ (http://ideone.com/iqHvqW). What compiler are you using and how? Variadic templates where added in c++11. — hmjd, May 15 '13 at 11:19
@RasmiRanjanNayak, that does not support variadic templates and only supports a subset of the c++11 features. — hmjd, May 15 '13 at 11:22

score 3 · Accepted Answer · edited May 23 '17 at 12:21

3

VS2010 does not support variadic templates. See C++11 Features. VS2012 does not support it either, according to that page, so upgrading is not a solution at the moment.

Search for c++03 mimic variadic templates to determine if there is an alternative (one example from this site: How to implement "Variadic Template" with pre-c++0x(VS2008)?).

edited May 23 '17 at 12:21

Community

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answered May 15 '13 at 11:29

hmjd

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score 1 · Answer 2 · answered May 15 '13 at 11:21

The problem is that you are using only the first, and not other template arguments. The g++ warning clearly explains it.

This example uses all arguments, and add a function for no arguments :

#include <iostream>

void fun()
{
    std::cout<<std::endl;
}
template<typename T, typename... Args>
void fun(T t, Args... args)
{
    std::cout << t;
    fun(args...);
}
int main()
{
    fun(1, 2.0, "Ranjan", "hi");//Error happens here
}

How to use template when number of arguments are not known?

2 Answers2