在下面的代码中,我希望函数globalFoo访问文件级别foo。怎么做呢?
class Foo {
foo() => print('Foo.foo');
globalFoo() => foo(); // How to call top level foo?
}
foo() => print('Global.foo');
main() {
final foo = new Foo();
foo.globalFoo();
}
您可以创建别名:
class Foo {
foo() => print('Foo.foo');
globalFoo() => _foo();
}
foo() => print('Global.foo');
final _foo = foo; // Alias foo with _foo
main() {
final foo = new Foo();
foo.globalFoo();
}