我如何 "hard-code" 检查 void 类型的相等性？

Question

如何“硬编码”检查 void 类型的相等性？

这段代码运行得很好

class Foo<T> {
  Type get ofType => T;
  bool isType<S>() => T == S;
}

final _foo = Foo<void>();

void main() {
  print(_foo.ofType);
  print(_foo.isType<void>());
}

但这甚至无法编译

class Foo<T> {
  Type get ofType => T;

  bool isTypeVoid() => T == void;
}

final _foo = Foo<void>();

void main() {
  print(_foo.ofType);
  print(_foo.isTypeVoid());
}

lib/main.dart:3:29:
Error: Expected an identifier, but got 'void'.
  bool isTypeVoid() => T == void;
                            ^^^^
lib/main.dart:3:33:
Error: Expected an identifier, but got ';'.
  bool isTypeVoid() => T == void;
                                ^
lib/main.dart:3:29:
Error: Expected ';' after this.
  bool isTypeVoid() => T == void;
                            ^^^^
lib/main.dart:3:29:
Error: Expected a class member, but got 'void'.
  bool isTypeVoid() => T == void;
                            ^^^^
Error: Compilation failed.

如果您想知道

，我需要在 assert() 中硬编码 T == void

assert(T == void)

但不起作用

Answer 1

似乎没有解决办法这是一些参考资料 [ 1, 2（第 216-217 页）]

至于这是一个答案，但不是解决方案我可以提供 nonsense 解决方法

class Nonsense<T> {
  const Nonsense();
  Type call() => T;
}

const _voidType = Nonsense<void>();

class Foo<T> {
  Type get ofType => T;
  bool isTypeVoid() => T == _voidType();
}

final _foo = Foo<void>();

void main() {
  print(_foo.ofType);
  print(_foo.isTypeVoid());
}

感谢 discord 上的灾难指导

我如何 "hard-code" 检查 void 类型的相等性？

how do I "hard-code" check equality for type void?

types

void

dart

flutter