在这种情况下，为什么需要指定默认构造对象的类型？

问：

我不明白为什么在下面我需要指定，而我没有：foobarstd::vector<int>{}foobar2

#include <iostream>
#include <memory>
#include <vector>
#include <tuple>

std::tuple<std::unique_ptr<int>, std::vector<int>> foobar() {
    std::unique_ptr<int> test = std::make_unique<int>(42);
    return { std::move(test), {} };    // <= this is a syntax error
    // return { std::move(test), std::vector<int>{} }  // <= this compiles...
}

std::tuple<int, std::vector<int>> foobar2() {
    return { {},  {} };
}

int main() {
    std::cout << *std::get<0>(foobar()) << "\n";
    std::cout << std::get<0>(foobar2()) << "\n";
    return 0;
}

来自 GCC 的错误消息是

<source>: In function 'std::tuple<std::unique_ptr<int, std::default_delete<int> >, std::vector<int, std::allocator<int> > > foobar()':
<source>:8:34: error: could not convert '{std::move<unique_ptr<int>&>(test), <brace-enclosed initializer list>()}' from '<brace-enclosed initializer list>' to 'std::tuple<std::unique_ptr<int, std::default_delete<int> >, std::vector<int, std::allocator<int> > >'
    8 |     return { std::move(test), {} };    // <= this is a syntax error
      |                                  ^
      |                                  |
      |                                  <brace-enclosed initializer list>
Compiler returned: 1