基于范围的for循环与const shared_ptr<> [英] range based for loop with const shared_ptr<>

问题描述

我有一个包含 shared_ptr<> 的容器，例如a 向量< shared_ptr< string>> v 并且我想迭代 v 表示常量。

此代码：

 矢量< shared_ptr< string> v; 
 v.push_back（make_shared< std :: string>（hallo））; 
 ... 
 
 for（const auto& s：v）{
 * s + =。 //<< == should be invalid 
} 
  

/ em>喜欢我想做的（指示 s 是 const ），但是当然不会是否有一个优雅的方法来遍历 shared_ptr 的容器。 const / code>这清楚表明内容不会被修改？

像

  for（shared_ptr< const string> s：v）{
 * s + =。 //<< == will not compile 
} 
  

不编译为其他原因：））

编辑：

错误。最初我是声明一个引用，导致编译器错误

  for（shared_ptr< const string>& s：v） {//<< ==不编译
 ... 
} 
  

如果你声明一个 shared_ptr< const string> 的例子工作。在我看来，这是一个很好的权衡，但这种方式指针被复制，这可能是耗时的循环与小代码和大容器。

解决方案

这是C ++的一个众所周知的限制，一些不认为是限制。

您要迭代 const ly，但不可变的指针并不意味着不可变的指针。

类型 shared_ptr< string> ; 和类型 shared_ptr< const string> 是无效的。

选项1

  for（const auto& ptr：v）{
 const auto& s = * ptr; 
 
 s + =。 //<< == is invalid 
} 
  

选项2

只是不要修改它。

I have a container with shared_ptr<>, e.g. a vector<shared_ptr<string>> v and I'd like to iterate over v indicating const-ness.

This code:

vector<shared_ptr<string>> v;
v.push_back(make_shared<std::string>("hallo"));
...

for (const auto &s : v) {
    *s += ".";   // <<== should be invalid
}

looks like what I want to do (indicating that s is const) but of course it does not make the string const.

Is there an elegant way to iterate over a container of shared_ptr which makes clear that the content won't be modified?

Something like

for (shared_ptr<const string> s : v) {
    *s += ".";   // <<== will not compile
}

(but this code would not compile for other reasons :))

Edit:

I made a mistake. Originally I was declaring a reference, which results in a compiler error

for (shared_ptr<const string> &s : v) {   // <<== does not compile
    ...
}

If you declare a shared_ptr<const string> the example works. In my eyes this is a good trade-off but this way the pointer gets copied which can be time consuming in loops with little code and big containers..

解决方案

This is a well-known limitation of C++ that some don't consider to be a limitation.

You want to iterate constly, but an immutable pointer doesn't imply an immutable pointee.

The type shared_ptr<string> and the type shared_ptr<const string> are effectively unrelated.

Option 1

for (const auto& ptr : v) {
    const auto& s = *ptr;

    s += ".";   // <<== is invalid
}

Option 2

Just don't modify it.

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