C++ template default parameter

Question

I know how simple templates and template specialization work, but I am stumped by this.

What is the T t = T() on the first line of the program doing? Is this a default parameter? And how does one determine the output of the program?

#include <iostream>

template<class T, T t = T()>
class A
{
private:
    template<bool b>
    class B
    {
    public:
        static const int m_n = b ? 1 : 0;
    };

public:
    static const int m_value = B<(t > T())>::m_n - B<(t < T())>::m_n;
};

int main()
{
    std::cout << A<int, -9>::m_value
              << A<bool, true>::m_value
              << A<char>::m_value << std::endl;

    return 0;
}

This is a question on a C++ assessment test that I am trying to understand.

Possible duplicate of *[Default template arguments for function templates](http://stackoverflow.com/questions/2447458/default-template-arguments-for-function-templates)*. — Peter Mortensen, Jun 23 '15 at 22:09

score 3 · Accepted Answer · edited Mar 07 '15 at 19:48

Yes. The second parameter is a default parameter for this template.

If you know this, the determination of the output should be fairly straight forward. I will do the first one for you:

A<int, -9>::m_value

int is the data type used for T, and the value of int t is -9.

This line:

static const int m_value = B<(t > T())>::m_n - B<(t < T())>::m_n;

Gets evaluated as this (where int() is zero):

static const int m_value = B<(-9 > 0)>::m_n - B<(-9 < 0)>::m_n;

Which evaluates as this:

static const int m_value = B<false>::m_n - B<true>::m_n;

Which evaluates as this:

static const int m_value = 0 - 1;

Which finally evaluates as this:

static const int m_value = -1;

So:

std::cout << A<int, -9>::m_value

Is the same as:

std::cout << -1

Now try figuring out the rest on your own.

thanks!. similarly bool() is 0, so the output of the above program is -110 — VarsMolta, Mar 07 '15 at 18:31

score 0 · Answer 2 · answered Mar 07 '15 at 18:15

0

Yes that is an example of default argument for class templates .You will find this example quite helpful https://msdn.microsoft.com/en-us/library/bys786s7.aspx

answered Mar 07 '15 at 18:15

Clyde D'Cruz

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Daniel Robertson · Answer 3 · 2015-03-07T18:37:56.893

In short, yes it does provide a default value for the second template parameter. You see the use of T t = T() in the line A<char>::m_value. Since the second template parameter is initialized to T() (the default constructor of T) by default t takes on the default value of whatever type you provide as the first template parameter. The program then compares the value given as the second template parameter with the default value of the type given as the first template parameter. Think of it like the following function, if I understand the class right, does the same thing.

template<class T>
// T t = T() in a function is the same as your T t = T()
// in your template parameters
int f(T t = T())
{
    return (T() == t) ? 0 : ((T() < t) ? 1 : -1);
}

Usage:

int main(int argc, char *argv[]) {
    std::cout << f<int(-9)
              << A<bool>(true)
              << A<char>() << std::endl;
}

The function returns 0 if t equals the default value of type T, -1 if t is less than the default value of type T, and +1 if t is greater than the default value of type T.

C++ template default parameter

3 Answers3