分配失败时，C ++ allocator :: allocate抛出还是返回nullptr？

问题描述

和不说 allocate 会做什么

当我使用标准分配器API编写容器时，应该

When I'm writing a container using standard allocator API, should I

1. 检查返回值并在 noexcept 版本成员函数中捕获异常（例如 push_back ，调整大小 ...）；

1. Check the return value and catch the exception in the noexcept version member function(E.g. push_back, resize...);

检查返回值并抛出异常，如果失败则抛出该异常

Check the return value and throw if fail in the exception-throwing one

这样，无论是否抛出异常得到正确的行为。

so that no matter it throws or not, I will get the correct behavior.

推荐答案

草稿n4659（对于C ++标准）说的是23.10.9。默认分配器[default.allocator]（强调我的）：

Draft n4659 for C++ standard says at 23.10.9 The default allocator [default.allocator] (emphasize mine):

T* allocate(size_t n);

2返回：指向大小为n * sizeof（T的存储数组的初始元素的指针），将
对齐为T类型的对象。

3备注：通过调用:: operator new（21.6.2）获得存储空间，但何时或$ b $未指定存储空间b该函数的调用频率。

4 抛出：如果无法获得存储，则抛出bad_alloc。

2 Returns: A pointer to the initial element of an array of storage of size n * sizeof(T), aligned appropriately for objects of type T.
3 Remarks: the storage is obtained by calling ::operator new (21.6.2), but it is unspecified when or how often this function is called.
4 Throws: bad_alloc if the storage cannot be obtained.

很明显，如果标准分配器无法分配存储，则会引发 bad_alloc 异常。

It makes it clear that the standard allocator will raise a bad_alloc exception if it cannot allocate storage.

以上是标准分配器。 20.5.3.5分配器要求[allocator.requirements]和表31 —分配器要求包含对任何分配器的要求：

Above is for the standard allocator. The requirement for any allocator are described in 20.5.3.5 Allocator requirements [allocator.requirements] and table 31 — Allocator requirements contains:

我的理解是分配仅在分配了内存后才能返回。因此，如果无法分配内存，分配器应该抛出一个适当的异常（即使非常合适也不一定要 bad_alloc ）。

My understanding is that allocate can return only when memory has been allocated. So the allocator should throw an appropriate exception (not necessarily bad_alloc even if it would be quite appropriate) if memory could not be allocated.

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