模板方法内的标准容器

2022-01-24 00:00:00 containers templates c++ std

您好.

我不太清楚如何解释自己，但我相信一段代码会让你明白我的意图:

I don't know very well how to explain myself, but I believe a piece of code will make you understand what I'm intenting to do :

template<class A, class B>
void myFunction(A<B>& list)
{
  typename A<B>::iterator current = list.begin();
  typename A<B>::iterator end = list.end();

  while (current != end)
  {
    current++;
  }
}

其中 A 是一个 STL 容器(向量、列表...).这就像 inception，但是有模板:一个模板，一个模板内，等等......

Where A is an STL container (vector, list...). It's like inception, but with templates : a template, inside a template, etc...

问题是:当您的模板的参数之一本身就是模板时，您会怎么做......并且仍然希望支持此模板支持的所有类型.

The thing is : what do you do when one of the params of your template is itself a template... and still want to support every types supported by this template.

这当然不能编译(它说A 不是模板").

This of course doesn't compile (it says 'A is not a template').

有人知道如何创建这样的模板吗?

Does someone knows how to create such a template ?

推荐答案

您正在寻找模板模板参数

You are looking for a template template parameter

template<template<class T, class All = std::allocator<T> > class A, class B>
void myFunction(A<B>& list)
{
  typename A<B>::iterator current = list.begin();
  typename A<B>::iterator end = list.end();

  while (current != end)
  {
    current++;
  }
}

但是，在您的特定情况下，我认为您最好只传递实例化容器，即

However, in your particular case, I think you'd be better off by just passing the intantiated container, that is, 

template<class C>
void myFunction(C& list)
{
   ...
}

这样使用

vector<char> v;
myFunction(v);

您的原始代码必须这样调用:

Your original code would have to be called like this:

myFunction<std::vector, char> (v)

更冗长，没有什么特别的好处

which is much more verbose and has no particular benefit

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