使用统一初始化语法从函数返回元组 [英] Returning a tuple from a function using uniform initialization syntax
问题描述
以下代码使用clang（libc ++）编译，并使用gcc（libstdc ++）失败。为什么gcc（libstdc ++）抱怨初始化列表？我认为返回参数使用统一的初始化语法。
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<double,double> dummy() {
return {2.0, 3.0};
}
int main() {
std::tuple<double,double> a = dummy();
return 0;
}
错误：第22行：转换为'std :: tuple'from initializer \
list将使用显式构造函数'constexpr std :: tuple< _T1，_T2> :: tuple（_U1& \
& _U2&& _U1 = double; _U2 = double; = void; _T\
1 = double;注意： GCC（libstdc ++）（和clang（libc ++））accept
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]’
/ p>
std::tuple<double,double> dummy {1.0, 2.0};
Isn't it the same case?不一样吗？
更新：：这是一个libc ++扩展，请参见 http://llvm.org/
推荐答案
不像 pair<> ，不可能遗漏构造元组。您必须使用 make_tuple（）：
#include <tuple>
std::tuple<double,double> dummy()
{
return std::make_tuple(2.0, 3.0); // OK
}
int main()
{
std::tuple<double,double> 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: std :: tuple 有一个可变参数构造函数，但它被标记为 explicit 。因此，它不能在这种情况下使用，其中临时必须是隐式可构造的。根据C ++ 11标准的第20.4.2节：
namespace std {
template <class... Types>
class tuple {
public:
[...]
explicit tuple(const Types&...); // Marked as explicit!
template <class... UTypes>
explicit tuple(UTypes&&...); // Marked as explicit!
For the same reason it is illegal to use copy-initialization syntax for initializing tuples:std::tuple<double, double> a = {1.0, 2.0}; // ERROR!
std::tuple<double, double> a{1.0, 2.0}; // OK
或者在将一个元组作为参数传递给函数时隐式地构造一个元组： p>
Or to construct a tuple implicitly when passing it as an argument to a function:
void f(std::tuple<double, double> t) { ... }
...
f({1.0, 2.0}); // ERROR!
f(make_tuple(1.0, 2.0)); // OK
因此，如果你构造你的 std :: tuple 显示在 dummy（）中返回时，不会发生编译错误：
Accordingly, if you construct your std::tuple explicitly when returning it in dummy(), no compilation error will occur:
#include <tuple>
std::tuple<double,double> dummy()
{
return std::tuple<double, double>{2.0, 3.0}; // OK
}
int main()
{
std::tuple<double,double> a = dummy();
return 0;
}
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