例如我有
private void test(Action<ValueTuple<string, int>> fn)
{
fn(("hello", 10));
}
test(t =>
{
var (s, i) = t;
Console.WriteLine(s);
Console.WriteLine(i);
});
我想写这样的东西
private void test(Action<ValueTuple<string, int>> fn)
{
fn(("hello", 10));
}
test((s,i) =>
{
Console.WriteLine(s);
Console.WriteLine(i);
});
这可以用一些适当的符号吗？
【问题讨论】：
您可以将其缩短为：
void test( Action<ValueTuple<string, int>> fn)
{
fn(("hello", 10));
}
test(((string s, int i) t) =>
{
Console.WriteLine(t.s);
Console.WriteLine(t.i);
});
希望有一天我们可以将元组中的参数分配到方法调用中：
void test(Action<ValueTuple<string, int>> fn)
{
fn(@("hello", 10)); // <-- made up syntax
}
test((s, i) =>
{
Console.WriteLine(s);
Console.WriteLine(i);
});
但目前没有。
【讨论】：
public static object[] ToArray<T1, T2>(this ValueTuple<T1, T2> t) { return new object[]{ t.Item1, t.Item2 }; })，您可以使用 fn.DynamicInvoke(("hello",10).ToArray()); - 当然，此时扩展 Invoke 可能更有意义。
() 之间指定参数，并在名称前加上类型。 ((string s, int i) t) => 在这方面与 `(string s) => 没有什么不同。
fn(@("hello", 10)); 没有意义。它仍在向函数传递一个元组，所以只需 fn(("hello", 10)); 就足够了。我们不需要任何特殊的语法，只需要更好的类型推断。想象一下最简单的test( ((str, inte)) =>，它将表明 (str,inte) 是一个未命名的元组类型参数。未命名参数允许将 str 和 inte 特殊处理为从传入元组解构的局部自动变量的名称。这与test( ((string, int)p) => 不同，其中参数是名称，(string,int) 是具有字段 Item1,Item2 的给定类型的非元组。
I. 带有 distinct-args 的 Action /Func 委托与 单 n 元组 参数的示例：
// 1. Action with 3 distinct 'int' parameters
Action<int, int, int> ArgsAction = (i1, i2, i3) => i1 += i2 += i3;
// 2. Func with 3 distinct 'int' parameters, returning 'long'
Func<int, int, int, long> ArgsFunc = (i1, i2, i3) => (long)i1 + i2 + i3;
// 3. Action with a single 3-tuple parameter
Action<(int, int, int)> TupleAction = args => args.Item1 += args.Item2 += args.Item3;
// 4. Action with a single 3-tuple parameter, returning 'long'
Func<(int, int, int), long> TupleFunc = args => (long)args.Item1 + args.Item2 + args.Item3;
II.演示以上示例的直接用法
long r;
// pass distinct params to multi-arg methods
ArgsAction(1, 2, 3); // 1.
r = ArgsFunc(1, 2, 3); // 2.
// pass tuple to tuple-taking methods
TupleAction((1, 2, 3)); // 3.
r = TupleFunc((1, 2, 3)); // 4.
接下来两节中的示例以各自的非本地参数形式调用委托。要延迟方法调用或为缓存或延迟/多次调用情况保留一个适应的委托，请参见 VI。和七。
III. 将元组分散（“splat”）到多参数方法中。
(1, 2, 3).Scatter(ArgsAction); // 1.
r = (1, 2, 3).Scatter(ArgsFunc); // 2.
IV.将不同的参数传递给元组获取方法：
TupleAction.Gather(1, 2, 3); // 3.
r = TupleFunc.Gather(1, 2, 3); // 4.
V.上面（III）和（IV）中使用的扩展方法Scatter和Gather：
// disperse n-tuple into Action arguments
public static void Scatter<T0, T1>(in this (T0 i0, T1 i1) t, Action<T0, T1> a) => a(t.i0, t.i1);
public static void Scatter<T0, T1, T2>(in this (T0 i0, T1 i1, T2 i2) t, Action<T0, T1, T2> a) => a(t.i0, t.i1, t.i2);
public static void Scatter<T0, T1, T2, T3>(in this (T0 i0, T1 i1, T2 i2, T3 i3) t, Action<T0, T1, T2, T3> a) => a(t.i0, t.i1, t.i2, t.i3);
// disperse n-tuple into Func arguments
public static TResult Scatter<T0, T1, TResult>(in this (T0 i0, T1 i1) t, Func<T0, T1, TResult> f) => f(t.i0, t.i1);
public static TResult Scatter<T0, T1, T2, TResult>(in this (T0 i0, T1 i1, T2 i2) t, Func<T0, T1, T2, TResult> f) => f(t.i0, t.i1, t.i2);
public static TResult Scatter<T0, T1, T2, T3, TResult>(in this (T0 i0, T1 i1, T2 i2, T3 i3) t, Func<T0, T1, T2, T3, TResult> f) => f(t.i0, t.i1, t.i2, t.i3);
// accumulate 'n' distinct args and pass into Action as an n-tuple
public static void Gather<T0, T1>(this Action<(T0, T1)> a, T0 i0, T1 i1) => a((i0, i1));
public static void Gather<T0, T1, T2>(this Action<(T0, T1, T2)> a, T0 i0, T1 i1, T2 i2) => a((i0, i1, i2));
public static void Gather<T0, T1, T2, T3>(this Action<(T0, T1, T2, T3)> a, T0 i0, T1 i1, T2 i2, T3 i3) => a((i0, i1, i2, i3));
// accumulate 'n' distinct args and pass into Func as an n-tuple
public static TResult Gather<T0, T1, TResult>(this Func<(T0, T1), TResult> f, T0 i0, T1 i1) => f((i0, i1));
public static TResult Gather<T0, T1, T2, TResult>(this Func<(T0, T1, T2), TResult> f, T0 i0, T1 i1, T2 i2) => f((i0, i1, i2));
public static TResult Gather<T0, T1, T2, T3, TResult>(this Func<(T0, T1, T2, T3), TResult> f, T0 i0, T1 i1, T2 i2, T3 i3) => f((i0, i1, i2, i3));
VI. 奖金回合。如果您计划以替代形式多次调用一个元组或不同参数的委托，或者如果您还没有准备好实际调用它，您可能希望显式地从 tuple 预转换委托- 采用 等效的 distinct-args 代表的形式，反之亦然。您可以缓存转换后的委托以供以后多次或任意重用。
var ga = ArgsAction.ToGathered(); // 1.
// later...
ga((1, 2, 3));
// ...
ga((4, 5, 6));
var gf = ArgsFunc.ToGathered(); // 2.
// later...
r = gf((1, 2, 3));
// ...
r = gf((4, 5, 6));
var sa = TupleAction.ToScattered(); // 3.
// later...
sa(1, 2, 3);
// ...
sa(4, 5, 6);
var sf = TupleFunc.ToScattered(); // 4.
// later...
r = sf(1, 2, 3);
// ...
r = sf(4, 5, 6);
// of course these approaches also supports in-situ usage:
ArgsAction.ToGathered()((1, 2, 3)); // 1.
r = ArgsFunc.ToGathered()((1, 2, 3)); // 2.
TupleAction.ToScattered()(1, 2, 3); // 3.
r = TupleFunc.ToScattered()(1, 2, 3); // 4.
VII.附赠示例的扩展方法见VI.
// convert tuple-taking Action delegate to distinct-args form
public static Action<T0, T1> ToScattered<T0, T1>(this Action<(T0, T1)> a) => (i0, i1) => a((i0, i1));
public static Action<T0, T1, T2> ToScattered<T0, T1, T2>(this Action<(T0, T1, T2)> a) => (i0, i1, i2) => a((i0, i1, i2));
public static Action<T0, T1, T2, T3> ToScattered<T0, T1, T2, T3>(this Action<(T0, T1, T2, T3)> a) => (i0, i1, i2, i3) => a((i0, i1, i2, i3));
// convert tuple-taking Func delegate to its distinct-args form
public static Func<T0, T1, TResult> ToScattered<T0, T1, TResult>(this Func<(T0, T1), TResult> f) => (i0, i1) => f((i0, i1));
public static Func<T0, T1, T2, TResult> ToScattered<T0, T1, T2, TResult>(this Func<(T0, T1, T2), TResult> f) => (i0, i1, i2) => f((i0, i1, i2));
public static Func<T0, T1, T2, T3, TResult> ToScattered<T0, T1, T2, T3, TResult>(this Func<(T0, T1, T2, T3), TResult> f) => (i0, i1, i2, i3) => f((i0, i1, i2, i3));
// convert distinct-args Action delegate to tuple-taking form
public static Action<(T0, T1)> ToGathered<T0, T1>(this Action<T0, T1> a) => t => a(t.Item1, t.Item2);
public static Action<(T0, T1, T2)> ToGathered<T0, T1, T2>(this Action<T0, T1, T2> a) => t => a(t.Item1, t.Item2, t.Item3);
public static Action<(T0, T1, T2, T3)> ToGathered<T0, T1, T2, T3>(this Action<T0, T1, T2, T3> a) => t => a(t.Item1, t.Item2, t.Item3, t.Item4);
// convert distinct-args Func delegate to its tuple-taking form
public static Func<(T0, T1), TResult> ToGathered<T0, T1, TResult>(this Func<T0, T1, TResult> f) => t => f(t.Item1, t.Item2);
public static Func<(T0, T1, T2), TResult> ToGathered<T0, T1, T2, TResult>(this Func<T0, T1, T2, TResult> f) => t => f(t.Item1, t.Item2, t.Item3);
public static Func<(T0, T1, T2, T3), TResult> ToGathered<T0, T1, T2, T3, TResult>(this Func<T0, T1, T2, T3, TResult> f) => t => f(t.Item1, t.Item2, t.Item3, t.Item4);
【讨论】：
有两种方法可以查看您的请求，但 C# 7.0 都不支持这两种方法。
M(tuple.first, tuple.second) 手动执行此操作。x => { var (first, second) = x; Write(first); Write(second); } 来手动执行此操作。csharplang design repo 中有一些提案正在讨论中。
【讨论】：
一种选择是使用 TupleSplatter (https://github.com/chartjunk/TupleSplatter)：
using TupleSplatter;
void test(Action<string, int> fn)
{
fn.SplatInvoke(("hello", 10));
// or
("hello", 10).Splat(fn);
}
test((s,i) => {
Console.WriteLine(s);
Console.WriteLine(i);
});
【讨论】：
这里有更简洁的语法变体，不需要任何额外的导入。不，它不能解决 cmets 中讨论的“飞溅”语法的愿望，但没有其他答案使用 ValueTuple 语法进行初始参数定义。
void test(Action<(string, int)> fn)
{
fn(("hello", 10));
}
// OR using optional named ValueTuple arguments
void test(Action<(string word, int num)> fn)
{
fn((word: "hello", num: 10));
}
使用 lambda 表达式的调用并不那么冗长，仍然可以使用最少的语法检索 ValueTuple 组件：
test( ((string, int) t) => {
var (s, i) = t;
Console.WriteLine(s);
Console.WriteLine(i);
});
【讨论】：
我能得到的最接近的。
public static class DeconstructExtensions
{
public static Action<T1, T2> Deconstruct<T1, T2>(this Action<(T1, T2)> action) => (a, b) => action((a, b));
public static Action<(T1, T2)> Construct<T1, T2>(this Action<T1, T2> action) => a => action(a.Item1, a.Item2);
}
class Test
{
private void fn((string, int) value) { }
private void test(Action<ValueTuple<string, int>> fn)
{
fn(("hello", 10));
}
private void Main()
{
var action = new Action<string, int>((s, i) =>
{
Console.WriteLine(s);
Console.WriteLine(i);
});
test(action.Construct());
}
}
【讨论】：
public static void CallBackTest( Action<(string Name, int Age)> fn)
{
fn(("MhamzaRajput", 23));
}
public static void Main()
{
CallBackTest((t) =>
{
var ( Name, Age ) = t;
Console.WriteLine(t.Name);
Console.WriteLine(t.Age);
});
}
输出
MhamzaRajput
23
【讨论】：