初始化对象时是否可以放弃放置新的返回值(Is it OK to discard placement new return value when initializing objects)

 这个问题来源于这个帖子中的评论部分，并且在那里也得到了答案。 但是，我认为仅仅留在评论部分是非常重要的。 所以我为此做了这个Q＆A。 
 
 Placement new可用于初始化分配的存储中的对象，例如， 
 
using vec_t = std::vector<int>;
auto p = (vec_t*)operator new(sizeof(vec_t));
new(p) vec_t{1, 2, 3}; // initialize a vec_t at p
 
 根据cppref ， 
 
 
 
 安置新的 
 
 
 如果提供了placement_params，它们将作为附加参数传递给分配函数。 在标准分配函数void* operator new(std::size_t, void*) ，这样的分配函数被称为“placement new”， 它只是返回其第二个参数不变 。 这用于在分配的存储中构造对象[...] 
 
 
 这意味着new(p) vec_t{1, 2, 3}只返回p ，而p = new(p) vec_t{1, 2, 3}看起来是多余的。 忽略返回值真的可以吗？ 

This question originates from the comment section in this thread, and has also got an answer there. However, I think it is too important to be left in the comment section only. So I made this Q&A for it.
 
Placement new can be used to initialize objects at allocated storage, e.g.,
 
using vec_t = std::vector<int>;
auto p = (vec_t*)operator new(sizeof(vec_t));
new(p) vec_t{1, 2, 3}; // initialize a vec_t at p
 
According to cppref,
 
 
 
Placement new
 
 
If placement_params are provided, they are passed to the allocation function as additional arguments. Such allocation functions are known as "placement new", after the standard allocation function void* operator new(std::size_t, void*), which simply returns its second argument unchanged. This is used to construct objects in allocated storage [...]
 
 
That means new(p) vec_t{1, 2, 3} simply returns p, and p = new(p) vec_t{1, 2, 3} looks redundant. Is it really OK to ignore the return value?

原文：https://stackoverflow.com/questions/49568858