std :: move与std :: shared_ptr在lambda中移动 [英] std::move with std::shared_ptr in lambda

问题描述

在lambdas中移动 std :: shared_ptr s时，我遇到一个奇怪的问题.我不确定这是否是错误，因为我可以使用g ++ v6.3和clang ++ v3.9进行复制.

I have a strange issue when moving std::shared_ptrs in lambdas. I am not sure if it is a bug or not, as I can reproduce with g++ v6.3 and clang++ v3.9.

当我编译并运行以下程序时:

When I compile and run the following program:

#include <iostream>
#include <memory>

void f(std::shared_ptr<int> ptr) {
  std::cout << 3 << " " << ptr.get() << std::endl;
}

int main() {
  auto ptr = std::make_shared<int>(42);
  std::cout << 1 << " " << ptr.get() << std::endl;
#ifdef LAMBDA
  auto lambda = [ptr]() {
#endif
    f(std::move(ptr));
    std::cout << 2 << " " << ptr.get() << std::endl;
#ifdef LAMBDA
  };
  lambda();
#endif
}

使用命令 c ++ -std = c ++ 14 -o test main.cpp&&./test 产生类似

1 0x55a49e601c30 1
3 0x55a49e601c30 1
2 0 0

但是，将编译命令更改为 c ++ -std = c ++ 14 -o test main.cpp -DLAMBDA 会使执行打印出我无法解释的内容:

However, changing the compilation command to c++ -std=c++14 -o test main.cpp -DLAMBDA makes the execution print something I cannot explain:

1 0x55a49e601c30 1
3 0x55a49e601c30 3
2 0x55a49e601c30 2

因此，看来 std :: move(move)在lambda中执行时，实际上并没有导致 shared_ptr 移动，也没有移动防止其参考计数器增加.

So, it appears that the std::move(move), when performed inside a lambda, does not actually cause the shared_ptr to be moved, nor does it prevent its reference counter to be incremented.

同样，我可以用clang ++和g ++复制它.

Again, I can reproduce this with clang++ and g++.

那怎么可能?

推荐答案

默认情况下，lambda中捕获的变量 ptr 是const，即 ptr 的类型为 const std :: shared_ptr< int> .

The captured variable ptr in a lambda is by default const, i.e. the type of ptr is const std::shared_ptr<int>.

std :: move 无法移出const对象，因此创建了副本.

std::move cannot move out const objects, so a copy is created instead.

如果您确实要移动它，则必须允许 ptr 是可变的:

If you really want to move it, the ptr must be allowed to be mutable:

auto lambda = [ptr]() mutable {
//                    ^~~~~~~

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