Expanding parameter pack containing initializer_list to constructor - c++

Question

I intend to use shared_ptr quite a bit in an upcoming project, so (not being aware of std::make_shared) I wanted to write a variadic template function spnew<T>(...) as a shared_ptr-returning stand-in for new. Everything went smoothly till I attempted to make use of a type whose constructor includes an initializer_list. I get the following from GCC 4.5.2 when I try to compile the minimal example below:
In function 'int main(int, char**)':
too many arguments to function 'std::shared_ptr spnew(Args ...) [with T = Example, Args = {}]'
In function 'std::shared_ptr spnew(Args ...) [with T = Example, Args = {}]':
no matching function for call to 'Example::Example()'
Oddly enough, I get equivalent errors if I substitute std::make_shared for spnew. In either case, it seems to be incorrectly deducing the parameters when an initializer_list is involved, erroneously treating Args... as empty. Here's the example:
#include <memory>
#include <string>
#include <vector>
struct Example {
// This constructor plays nice.
Example(const char* t, const char* c) :
title(t), contents(1, c) {}
// This one does not.
Example(const char* t, std::initializer_list<const char*> c) :
title(t), contents(c.begin(), c.end()) {}
std::string title;
std::vector<std::string> contents;
};
// This ought to be trivial.
template<class T, class... Args>
std::shared_ptr<T> spnew(Args... args) {
return std::shared_ptr<T>(new T(args...));
}
// And here are the test cases, which don't interfere with one another.
int main(int argc, char** argv) {
auto succeeds = spnew<Example>("foo", "bar");
auto fails = spnew<Example>("foo", {"bar"});
}
Is this just an oversight on my part, or a bug?

Answer 1

You could do this -
#include <memory>
#include <string>
#include <iostream>
#include <vector>
struct Example {
template<class... Args>
Example(const char* t, Args... tail) : title(t)
{
Build(tail...);
}
template<class T, class... Args>
void Build(T head, Args... tail)
{
contents.push_back(std::string(head));
Build(tail...);
}
template<class T>
void Build(T head)
{
contents.push_back(std::string(head));
}
void Build() {}
std::string title;
std::vector<std::string> contents;
};
template<class T, class... Args>
std::shared_ptr<T> spnew(Args... args) {
return std::shared_ptr<T>(new T(args...));
}
int main(int argc, char** argv) {
auto succeeds = spnew<Example>("foo", "bar");
auto fails = spnew<Example>("foo", "bar", "poo", "doo");
std::cout << "succeeds->contents contains..." << std::endl;
for ( auto s : succeeds->contents ) std::cout << s << std::endl;
std::cout << std::endl << "fails->contents contains..." << std::endl;
for ( auto s : fails->contents ) std::cout << s << std::endl;
}
This, despite the generic templates is type safe as the compiler will complain about
the contents.push_back if the passed type is not convertible to a const char *.
As described above, your code was working fine with gcc 4.6 however the warning you get is explained here
why-doesnt-my-template-accept-an-initializer-list, and is possibly not standards
compliant, although the c++0x standard is yet to be published so this could change.

Answer 2

With gcc-4.7 (probably would work on gcc-4.6 too, just branched) with warnings:
foo.cpp: In function ‘int main(int, char**)’:
foo.cpp:29:47: warning: deducing ‘Args ...’ as ‘std::initializer_list<const
char*>’ [enabled by default]
foo.cpp:22:20: warning: in call to ‘std::shared_ptr<_Tp1> spnew(Args ...)
[with T = Example, Args = {const char*, std::initializer_list<const
char*>}]’ [enabled by default]
foo.cpp:29:47: warning: (you can disable this with -fno-deduce-init-list)
[enabled by default]
I'm not sure why anyone would want to beef about init-list deduction though.
There is a related thread:
Why doesn't my template accept an initializer list
Basically, a bare init-list doesn't have a type.