是否可以从集合中提取包含唯一/共享指针的字段的结构？

Question

从本质上说，我有一组struct的实例，我想提取一个实例，修改字段。其中一个字段是唯一的ptr。我不太擅长读取c++错误，在提取时，该字段似乎被删除了。即调用唯一指针的析构函数。

在这里，当我试图访问字段“权重”时

我得到了这样的信息：

SUMMARY: AddressSanitizer: heap-use-after-free /usr/lib/llvm-11/include/c++/v1/memory:2490:19 in std::__1::unique_ptr<int, std::__1::default_delete<int> >::get() const
[ctest] Shadow bytes around the buggy address:
[ctest]   0x0c0e7fff84a0: fd fd fd fd fd fa fa fa fa fa fd fd fd fd fd fd
[ctest]   0x0c0e7fff84b0: fd fd fd fa fa fa fa fa fd fd fd fd fd fd fd fd
[ctest]   0x0c0e7fff84c0: fd fa fa fa fa fa fd fd fd fd fd fd fd fd fd fa
[ctest]   0x0c0e7fff84d0: fa fa fa fa 00 00 00 00 00 00 00 00 00 fa fa fa
[ctest]   0x0c0e7fff84e0: fa fa 00 00 00 00 00 00 00 00 00 fa fa fa fa fa
[ctest] =>0x0c0e7fff84f0: fd fd fd fd fd fd fd fd[fd]fa fa fa fa fa 00 00
[ctest]   0x0c0e7fff8500: 00 00 00 00 00 00 00 fa fa fa fa fa fa fa fa fa
[ctest]   0x0c0e7fff8510: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
[ctest]   0x0c0e7fff8520: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
[ctest]   0x0c0e7fff8530: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
[ctest]   0x0c0e7fff8540: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
[ctest] Shadow byte legend (one shadow byte represents 8 application bytes):
[ctest]   Addressable:           00
[ctest]   Partially addressable: 01 02 03 04 05 06 07 
[ctest]   Heap left redzone:       fa
[ctest]   Freed heap region:       fd
[ctest]   Stack left redzone:      f1
[ctest]   Stack mid redzone:       f2
[ctest]   Stack right redzone:     f3
[ctest]   Stack after return:      f5
[ctest]   Stack use after scope:   f8
[ctest]   Global redzone:          f9
[ctest]   Global init order:       f6
[ctest]   Poisoned by user:        f7
[ctest]   Container overflow:      fc
[ctest]   Array cookie:            ac
[ctest]   Intra object redzone:    bb
[ctest]   ASan internal:           fe
[ctest]   Left alloca redzone:     ca
[ctest]   Right alloca redzone:    cb
[ctest]   Shadow gap:              cc

Struct A {
    std::unique_ptr<int> weight;
}

auto& element = set_.extract(set_iterator).value();
std::cout << *(element.weight);

Answer 1

该节点在提取时不会被删除，而是在您使用它之前不久删除。

具体来说，您不保存节点句柄.extract()返回，而是直接获得它管理的部分节点的引用。

在完整语句的末尾，该引用会变得悬空，因为没有人保持节点句柄是活动的。

因此，AddressSanitizer检测堆后使用-无.

存储节点句柄，或保存所需节点的部分，但避免挂起引用。