F.42: Return a T* to indicate a position (only)
F.42：T*返回值应该只用来指明位置
Reason（原因）
That's what pointers are good for. Returning a T* to transfer ownership is a misuse.
这是指针的强项。使用T*返回所有权是错误的用法。
Example（示例）
Node* find(Node* t, const string& s) // find s in a binary tree of Nodes{ if (!t || t->name == s) return t; if ((auto p = find(t->left, s))) return p; if ((auto p = find(t->right, s))) return p; return nullptr;}If it isn't the nullptr, the pointer returned by find indicates a Node holding s. Importantly, that does not imply a transfer of ownership of the pointed-to object to the caller.
如果返回值不为nullptr，则find返回的指针表示持有s的Node。重要的是，它不包含向调用者传递指针所指向的对象的所有权的含义。
Note（注意）
Positions can also be transferred by iterators, indices, and references. A reference is often a superior alternative to a pointer if there is no need to use nullptr or if the object referred to should not change.
位置也可以通过迭代器，索引和引用传递。如果不需要使用nullptr或者对象不希望被修改，引用通常是比指针更好的选择。
Note（注意）
Do not return a pointer to something that is not in the caller's scope; see F.43.
不要返回指向不属于调用者范围的某物的指针。
See also: discussion of dangling pointer prevention
参见：关于防止悬空指针的讨论。
Enforcement（实施建议）
- Flag
delete, std::free(), etc. applied to a plain T*. Only owners should be deleted.
标示运用在直接指针上的delete，std::free()等操作。只用所有者才可以删除。 - Flag
new, malloc(), etc. assigned to a plain T*. Only owners should be responsible for deletion.
标记将结果分配个直接指针的new，malloc()等处理。只有所有者才有销毁的责任。
阅读更多更新文章，请关注微信公众号【面向对象思考】
