模板约束内的模式匹配 [英] Pattern-matching inside a template constraint

问题描述

该问题基于安德烈（Andrei）对我对签名约束的问题的回答。

This question is based on Andrei's answer to my question on signature constraints.

struct S(int x, int y) {
  void fun(T)(T t) if (is(T U == S!(a, b), int a, int b)) { }
}

template s(int a, int b) {
  enum result = S!(a,b)();
  alias result s;
}

void main() {

  auto s1 = S!(1, 1)();
  auto s2 = S!(2, 2)();
  auto s3 = s!(3, 3);
  auto s4 = s!(4, 4);

  s1.fun(s1);  // ok
  s1.fun(s2);  // ok
  s1.fun(s3);  // compile error
  s3.fun(s1);  // ok
  s3.fun(s3);  // compile error
  s3.fun(s4);  // compile error
}

我不明白为什么代码会产生编译错误。有想法吗？

I don't understand why the code is producing compile errors. Any ideas?

推荐答案

首先，我不建议使用裸模板来生成对象/结构的实例，因为您本质上要求对象是可CTFE的。如果需要实例，最好的选择是从模板化函数中返回：

First, I wouldn't recommend using a naked template to generate an instance of an object/struct, because you're essentially requiring the object to be CTFE-able. If you need an instance your best option is to return it from a templated function:

@property S!(a, b) s(int a, int b)()
{
    return S!(a, b)();
}

但是，这似乎仍然无法与模板约束一起使用。我认为这必须是前端错误。据我所知，似乎返回的类型不能在is（）表达式中正确检查，除非它已在其他地方实例化，例如：

However this still doesn't seem to work with the template constraint. I think this has to be a front-end bug. From what I can tell it seems that the type returned cannot be properly checked in the is() expression unless it was already instantiated somewhere else, e.g.:

struct S(int x, int y) 
{
    void fun(T)(T t) 
        if (is(T U == S!(a, b), int a, int b))
    {

    }
}

@property S!(a, b) s(int a, int b)()
{
    return S!(a, b)();
}

void main() 
{
    auto s1 = S!(1, 1)();
    auto s2 = S!(2, 2)();  // comment out and you get errors in fun() call
    auto s3 = s!(2, 2);
    s1.fun(s3);
}

我将其记录为错误。

编辑：归档为问题8493 。

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