The following code compiles with clang (libc++) and fails with gcc (libstdc++). Why does gcc (libstdc++) complains about an initializer list? I thought the return argument was using uniform initialization syntax.
std::tuple dummy() {
return {2.0, 3.0};
}
int main() {
std::tuple a = dummy();
return 0;
}
Error: line 22: converting to ‘std::tuple’ from initializer \
list would use explicit constructor ‘constexpr std::tuple<_T1, _T2>::tuple(_U1&\
&, _U2&&) [with _U1 = double; _U2 = double; = void; _T\
1 = double; _T2 = double]’
Note: GCC (libstdc++) (and clang (libc++)) accept
std::tuple dummy {1.0, 2.0};
Isn't it the same case?
Update: this is a libc++ extension, see http://llvm.org/bugs/show_bug.cgi?id=15299 and also answer by Howard Hinnant below.
Answer
Unlike for pair<>
, implicit construction of a tuple<>
is not possible unfortunately. You have to use make_tuple()
:
#include
std::tuple dummy()
{
return std::make_tuple(2.0, 3.0); // OK
}
int main()
{
std::tuple a = dummy();
return 0;
}
std::tuple
has a variadic constructor, but it is marked as explicit
. Thus, it cannot be used in this situation, where a temporary must be implicitly constructible. Per Paragraph 20.4.2 of the C++11 Standard:
namespace std {
template
class tuple {
public:
[...]
explicit tuple(const Types&...); // Marked as explicit!
template
explicit tuple(UTypes&&...); // Marked as explicit!
For the same reason it is illegal to use copy-initialization syntax for initializing tuples:
std::tuple a = {1.0, 2.0}; // ERROR!
std::tuple a{1.0, 2.0}; // OK
Or to construct a tuple implicitly when passing it as an argument to a function:
void f(std::tuple t) { ... }
...
f({1.0, 2.0}); // ERROR!
f(make_tuple(1.0, 2.0)); // OK
Accordingly, if you construct your std::tuple
explicitly when returning it in dummy()
, no compilation error will occur:
#include
std::tuple dummy()
{
return std::tuple{2.0, 3.0}; // OK
}
int main()
{
std::tuple a = dummy();
return 0;
}
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