与std :: shared_ptr和std :: unique_ptr相比，std :: optional有什么优势? [英] What's the advantage of `std::optional` over `std::shared_ptr` and `std::unique_ptr`?

问题描述

std::optional的理由是言而明达，它可能或可能不包含值.因此，如果我们不需要它，则可以省去构造一个可能很大的对象的工作.

The reasoning of std::optional is made by saying that it may or may not contain a value. Hence, it saves us the effort of constructing a, probably, big object, if we don't need it.

例如，这里的工厂，不会在某些情况下构造对象条件不满足:

For example, a factory here, will not construt the object if some condition is not met:

#include <string>
#include <iostream>
#include <optional>

std::optional<std::string> create(bool b) 
{
    if(b)
        return "Godzilla"; //string is constructed
    else
        return {}; //no construction of the string required
}

但是，这与此有何不同?

But then how is this different from this:

std::shared_ptr<std::string> create(bool b) 
{
    if(b)
        return std::make_shared<std::string>("Godzilla"); //string is constructed
    else
        return nullptr; //no construction of the string required
}

通过在通常情况下仅使用std::shared_ptr来添加std::optional，我们能赢得什么?

What is it that we win by adding std::optional over just using std::shared_ptr in general?

推荐答案

通过添加std :: optional而不是仅使用std :: shared_ptr来赢得什么呢?

What is it that we win by adding std::optional over just using std::shared_ptr in general?

比方说，您需要从带有标志"not a value"的函数返回符号.如果您将std::shared_ptr用于此用途，则会产生巨大的开销-char将分配到动态内存中，而std::shared_ptr将保留控制块.而另一边 std :: optional :

Let's say you need to return a symbol from a function with flag "not a value". If you would use std::shared_ptr for that you would have huge overhead - char would be allocated in dynamic memory, plus std::shared_ptr would maintain control block. While std::optional on another side:

如果可选包含值，则保证该值为 分配为可选对象足迹的一部分，即无动态 内存分配曾经发生过.因此，可选对象对 对象，而不是指针，即使operator *()和operator->() 被定义.

If an optional contains a value, the value is guaranteed to be allocated as part of the optional object footprint, i.e. no dynamic memory allocation ever takes place. Thus, an optional object models an object, not a pointer, even though the operator*() and operator->() are defined.

因此不涉及动态内存分配，即使与原始指针相比，差异也很明显.

so no dynamic memory allocation is involved an difference comparing even to the raw pointer could be significant.

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