c++ - 将 unique_ptr 移入 lambda 时，为什么不能调用 reset？

Question

当将 std::unique_ptr 移动到 lambda 中时，无法在其上调用 reset()，因为它似乎是 const :

error C2662: void std::unique_ptr<int,std::default_delete<_Ty>>::reset(int *) noexcept': cannot convert 'this' pointer from 'const std::unique_ptr<int,std::default_delete<_Ty>>' to 'std::unique_ptr<int,std::default_delete<_Ty>> &

#include <memory>
int main()
{
    auto u = std::unique_ptr<int>();
    auto l = [v = std::move(u)]{
        v.reset(); // this doesn't compile
    };
}

1. 为什么会这样？
2. 是否有可能以另一种方式捕获 std::unique_ptr 以允许在 lambda 中调用 reset()(使用 C++17 或更高版本)？

Answer 1

1. Why does this happen?

因为 lambda 的函数调用运算符,

Unless the keyword mutable was used in the lambda-expression, the function-call operator is const-qualified and the objects that were captured by copy are non-modifiable from inside this operator().

和

2. Is it possible to capture the std::unique_ptr in another way which allows to call reset() within the lambda

你需要标记它可变。

mutable: allows body to modify the parameters captured by copy, and to call their non-const member functions

例如

auto l = [v = std::move(u)]() mutable {
    v.reset();
};