将变量作为模板参数传递 [英] Passing a variable as a template argument

问题描述

我正在使用一个公开使用接口的库。该库的功能之一是这样的：

I am working with a library which exposes an interface to work with. One of the functions of this library is like this :

template <int a>
void modify(){}

我必须将参数从1修改为10，即调用 modify 的模板参数为1到10。为此，我编写了这段代码（基本版本的代码，实际代码要大得多）。

I have to modify parameters from 1 to 10 i.e. call modify with with template arguments from 1 to 10. For that I wrote this code (a basic version of code, actual code is much larger).

for(int i=0; i<10; i++){
    modify<i>();
}

在编译时收到以下错误

error: 'i' cannot appear in constant-expression

浏览了Internet上的某些链接后，我知道我不能将任何值作为模板参数传递，而该值在编译时未进行评估。
我的问题如下：
1.为什么编译器在编译时不能评估 i ？
2.在不更改API接口的情况下，还有其他目标可以实现吗？

After going through some links on the internet, I came to know that I cannot pass any value as template argument which is not evaluated at compile time. My question are as follows: 1. Why can't compiler evaluate i at compile time? 2. Is there any other to achieve the objective I am trying to achieve without changing the API interface?

还有另一件事我想做。调用Modify作为Modify，其中VAR是某些函数计算的输出。我怎样才能做到这一点？

There is another thing I want to do. Call modify as modify where VAR is the output of some functional computation. How can I do that?

推荐答案

在编译时 i 的值是什么（不是常数）？除非执行循环，否则无法回答。但是执行不是编译
既然没有答案，编译器就不能这样做。

What is the value of i (that is not a constant) at compile time? There is no way to answer unless executing the loop. But executing is not "compiling" Since there is no answer, the compiler cannot do that.

模板不是要执行的算法，而是要执行的宏扩展生成代码。
您可以做的是依靠专业化来实现递归迭代，例如：

Templates are not algorithm to be executed, but macros to be expanded to produce code. What you can do is rely on specialization to implement iteration by recursion, like here:

#include <iostream>

template<int i>
void modify()
{ std::cout << "modify<"<<i<<">"<< std::endl; }

template<int x, int to>
struct static_for
{
    void operator()() 
    {  modify<x>();  static_for<x+1,to>()(); }
};

template<int to>
struct static_for<to,to>
{
    void operator()() 
    {}
};


int main()
{
    static_for<0,10>()();
}

请注意，这样做实际上是实例化了10个函数名为
modify< 0> ... modify< 9> ，分别由<$ c $调用c> static_for< 0,10> :: operator（） ... static_for< 9,10> :: operator（）。

Note that, by doing this, you are, in fact, instantiating 10 functions named modify<0> ... modify<9>, called respectively by static_for<0,10>::operator() ... static_for<9,10>::operator().

迭代将结束，因为 static_for< 10,10> 将从采用两个相同值的专业化实例化，没什么。

The iteration ends because static_for<10,10> will be instantiated from the specialization that takes two identical values, that does nothing.

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