Function with parameter pack with sizeof ... (args) == 0 as base case doesn't compile

Question

Here's the code of my function:

#include <iostream>
#include <type_traits>
#include <algorithm>

template <typename Head, typename ... Args>
std::common_type_t<Head, Args...> mx(Head n, Args ... args)
{
    if (sizeof ... (args) == 0)
        return n;
    else
        return std::max(n, mx(args ...));
}

int main()
{
    std::cout << mx(3, 4, 5);
}

I got compile errors:

main.cpp: In instantiation of 'std::common_type_t<Head, Args ...> mx(Head, Args ...) [with Head = int; Args = {}; std::common_type_t<Head, Args ...> = int]': main.cpp:11:24:
recursively required from 'std::common_type_t<Head, Args ...> mx(Head, Args ...) [with Head = int; Args = {int}; std::common_type_t<Head, Args ...> = int]' main.cpp:11:24: required from 'std::common_type_t<Head, Args ...> mx(Head, Args ...) [with Head = int; Args = {int, int}; std::common_type_t<Head, Args ...> = int]' main.cpp:16:25: required from here main.cpp:11:24: error: no matching function for call to 'mx()' 11 | return std::max(n, mx(args ...)); | ~~^~~~~~~~~~ main.cpp:6:35: note: candidate: 'template<class Head, class ... Args> std::common_type_t<Head, Args ...> mx(Head, Args ...)' 6 | std::common_type_t<Head, Args...> mx(Head n, Args ... args) | ^~ main.cpp:6:35: note: template argument deduction/substitution failed: main.cpp:11:24: note: candidate expects at least 1 argument, 0 provided 11 | return std::max(n, mx(args ...)); | ~~^~~~~~~~~~

Of course I can write this more properly, like this:

template <typename Head>
std::common_type_t<Head> mx(Head n)
{
    return n;
}

template <typename Head, typename ... Args>
std::common_type_t<Head, Args...> mx(Head n, Args ... args)
{
    return std::max(n, mx(args ...));
}

But still, I don't understand why my first option doesn't work. Judging by errors, it somehow tries to call recursive version of function even if there's no arguments in parameter pack. But that does not make any sense to me since I considered this case. What's the problem and can I fix it?

score 5 · Accepted Answer · answered Dec 18 '21 at 20:42

5

Even if

if (sizeof ... (args) == 0)

the entire function must be well-formed C++.

    return std::max(n, mx(args ...));

This still must be valid C++, even if won't get executed. If, outside of template context, you have an if (1), the else part must still be valid C++, you can't just throw randomly-generated gibberish in there, and this is the same thing. And when sizeof...(args) is 0 the function call becomes mx() and that, of course, has no valid overload.

What you want to do, instead, is use if constexpr instead of your garden-variety if.

if constexpr (sizeof ... (args) == 0)

answered Dec 18 '21 at 20:42

Sam Varshavchik

114,536
5
94
148

I see now, when there's no arguments in parameter pack I still somehow try to call function without arguments. I mean, I'm not trying to do it but still there's is some piece of code that wants to do that, even if that piece of code will never executes. – Learpcs Dec 18 '21 at 20:49
You don't "try to call function", you're still "trying to compile the function". There's a big difference between compiling code and executing code. Your C++ library, that you linked to your C++ program, will have a huge pile of code that your C++ program will never execute. But it still has to get compiled, at some point. – Sam Varshavchik Dec 18 '21 at 20:49

Function with parameter pack with sizeof ... (args) == 0 as base case doesn't compile

1 Answers1