此memcpy是否将所有数据正确发送到我分配给堆的heapNode?即在此过程中没有任何损坏&*数据是否完整?

Does this memcpy all data correctly to my heap allocated heapNode? i.e. nothing is corrupted in the process & the *data is intact?

这个问题的答案是， memcpy 使用 shallow copy 概念.在 shallow copy 中，原始结构(节点)中的指针 TSNode node = get_node(); 将被复制到新节点 heapNode ( TSNode * heapNode )一点一点地调用逐位复制.因此，新的 heapNode 指针也将指向与原始节点的指针( TSNode node = get_node(); )指针相同的位置，该指针是通过复制值的位置memcpy .因此，一旦控件从函数 myFun 返回，则原始节点将从内存中释放.因此，您的新节点指针将变为悬挂指针.这是 memcpy 的严重副作用.

The answer to this question is, memcpy use shallow copy concept. In shallow copy the pointer in your original structure (node) TSNode node = get_node(); will be copied to the new node heapNode (TSNode* heapNode) bit by bit also call bit wise copy. So your new heapNode pointers will also pointing to the same location as your original node's (TSNode node = get_node();) pointers from where you copied the value with memcpy. Hence once control returns from function myFun your original node will be released from memory. Therefore your new node pointers will become dangling pointers. That is a critical side effect of memcpy.