使用std :: optional标准化，我们可以停止在新代码中使用nullptr并弃用它吗? [英] With std::optional standardizing, can we stop using nullptr in new code and deprecate it?

问题描述

从远古时代开始，当将指针传递给函数或从函数传递指针时，我们倾向于对空指针进行特殊处理:

From time immemorial, when passing pointers to or from functions, we tend to special-case the null pointer:

 p = get_pointer_to_the_foo(args);
 if (p == nullptr) { /* foo is inaccessible, do one thing */ }
 else { /* we can access foo, do something else */ }

，这是从C继承的.现在，我们偶尔会对其他类型执行相同的操作，例如使用带符号的类型表示有效的非负值，例如-1或表示错误.

and this is an inheritance from C. Now, we occasionally would do the same with other types, e.g. using a signed type to represent either a valid non-negative value, and, say, -1 or to indicate an error.

现在将开始使用std::optional最终弃用后一种模式:std::optional<unsigned>是nullopt或非负值.但是-指针呢?毕竟，nullptr只是无数无效指针值之一.因此，在编写新代码时(全部都是C ++ 17时)-我们是否应该从本质上忘记它，而绕过std::optional<foo_t*>或假定非空foo_t *呢?

The latter pattern will now be finally deprecated with the onset of std::optional: std::optional<unsigned> is either nullopt or a non-negative value. But - what about pointers? After all, nullptr is just one of innumerable invalid pointer values. So, when writing new code (when all of it is C++17) - should we essentially forget about it, and pass around either std::optional<foo_t*>'s or assumed-non-null foo_t *'s?

推荐答案

Pointer已经嵌入了可选语义.因此std::optional<T*>至少可以说是多余的，并且该功能的引入不会对很多原始指针产生影响.

Pointer embed the optional semantic already. Therefore std::optional<T*> would be redundant to say the least and the introduction of said feature won't impact much raw pointers.

对于std::unique_ptr，std::shared_ptr和std::weak_ptr也是如此，所有这些也可以是nullptr.

This is also true with std::unique_ptr, std::shared_ptr and std::weak_ptr all of which can also be nullptr.

就像使用任何新功能一样，我们必须思考什么是实用的，而不是我们天生的在可行的地方使用它的冲动.

Just like with any new feature we have to ponder what's pragmatic versus our innate urge to use it everywhere it's feasible.

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