为什么 `polymorphic_allocator` 采用 `memory_resource` 指针而不是引用？
Why does `polymorphic_allocator` take a `memory_resource` pointer and not a reference?
C++17 标准说：
polymorphic_allocator(memory_resource* r);
Requires: r is non-null.
Effects: Sets memory_rsrc to r.
Throws: Nothing.
[ Note: This constructor provides an implicit conversion from memory_resource*. — end note ]
如果 "requires" 子句提到 r 必须接受 memory_resource* 而不是 memory_resource& 有什么意义非空？
Bloomberg¹ 风格指南鼓励接受将被指针而不是引用改变的参数，以便调用方的 & 符号成为改变的视觉标记。但是，标准中没有这样的先例。
为什么r被当作指针而不是引用？
¹ pmr 在 Bloomberg 的大力参与下实现了标准化，因为该公司使用多态分配器模型。
Note that the memory-resource library is designed so that the ShoppingList constructor accepts a pointer to a memory_resource rather than a reference to a memory_resource. It was noted that one common practice is to use references rather than pointers in situations where a null pointer is out of contract. However, there is a more compelling practice of avoiding constructors that take objects by reference and store their addresses. We also want to avoid passing non-const references, as that, too, is usually considered bad practice (except in overloaded operators).
C++17 标准说：
polymorphic_allocator(memory_resource* r);
Requires:
ris non-null.Effects: Sets
memory_rsrctor.Throws: Nothing.
[ Note: This constructor provides an implicit conversion from
memory_resource*. — end note ]
如果 "requires" 子句提到 r 必须接受 memory_resource* 而不是 memory_resource& 有什么意义非空？
Bloomberg¹ 风格指南鼓励接受将被指针而不是引用改变的参数，以便调用方的 & 符号成为改变的视觉标记。但是，标准中没有这样的先例。
为什么r被当作指针而不是引用？
¹ pmr 在 Bloomberg 的大力参与下实现了标准化，因为该公司使用多态分配器模型。
Note that the memory-resource library is designed so that the
ShoppingListconstructor accepts a pointer to amemory_resourcerather than a reference to amemory_resource. It was noted that one common practice is to use references rather than pointers in situations where a null pointer is out of contract. However, there is a more compelling practice of avoiding constructors that take objects by reference and store their addresses. We also want to avoid passing non-const references, as that, too, is usually considered bad practice (except in overloaded operators).