需要帮助了解lambda(currying)

Question

我正在阅读Accelerated C#我真的不明白以下代码:

public static Func<TArg1, TResult> Bind2nd<TArg1, TArg2, TResult> (
    this Func<TArg1, TArg2, TResult> func,
    TArg2 constant ) 
{
    return (x) => func( x, constant );
}

在最后一行x指的是什么？还有另一个:

public static Func<TArg2, Func<TArg1, TResult>> Bind2nd<TArg1, TArg2, TResult>
( this Func<TArg1, TArg2, TResult> func )
{
    return (y) => (x) => func( x, y );
}

我该如何评价这个？(y) => (x) => func( x, y )通过哪里......它确实令人困惑.

Answer 1

我们先简化一下代码:

Func<int, int> B(Func<int, int, int> f, int c)
{
    return x=>f(x, c);
}

这与以下内容相同:

class Locals
{
    public int c;
    public Func<int, int, int> f;
    public int Magic(int x) { return f(x, c); }
}
Func<int, int> B(Func<int, int, int> f, int c)
{
    Locals locals = new Locals();
    locals.f = f;
    locals.c = c;
    return locals.Magic;
}

现在很清楚x指的是什么？x是"魔术"功能的参数.

现在你可以像这样使用B:

Func<int, int, int> adder = (a, b)=>a+b;
Func<int, int> addTen = B(adder, 10);
int thirty = addTen(20);

合理？看看这里发生了什么？我们正在使用两个参数的函数并将其中一个参数"固定"为常量.所以它成为一个参数的函数.

第二个例子更进了一步.再次,简化以摆脱困境,以便您可以更容易地理解它:

Func<int, Func<int, int>> B2(Func<int, int, int> f) 
{
    return y=>x=>f(x,y);
}

这是一样的

class Locals3
{
    public int y;
    public int Magic3(int x)
    {
       return x + this.y;
    }
}
class Locals2
{
    public Func<int, int, int> f;
    public Func<int, int> Magic2(int y)
    {
        Locals3 locals = new Locals3;
        locals.y = y;
        return locals.Magic3;
    }
}

Func<int, Func<int, int>> B2(Func<int, int, int> f) 
{
    Locals2 locals = new Locals2();
    locals.f = f;   
    return locals.Magic2;
}

所以你说

Func<int, int, int> adder = (a, b)=>a+b;
Func<int, Func<int, int>> makeFixedAdder = B2(adder);
Func<int, int> add10 = makeFixedAdder(10);
int thirty = add10(20);

B是参数修复器.B2 为您制作参数修复工具.

然而,这不是点 B2的.B2的重点是:

adder(20, 10);

给出了相同的结果

B2(adder)(20)(10)

B2 将两个参数的一个函数转换为每个参数的两个函数.

合理？