julia - 如何将一个函数作为参数传递给 Julia 中的另一个函数？

Question

我们可以将一个函数作为 julia 中另一个函数的参数传递吗？它是如何工作的？这是否意味着输入函数在调用函数之前运行，或者输入函数只有在调用函数专门调用它时才会被调用？

Answer 1

Julia 最大的特点之一是您可以自己挖掘出这些问题的答案。与其做实验，然后观察表面上的行为，直接问 Julia 它在幕后做了什么，是一种更方便、更简洁的方式来获得答案。让我们从其他两个答案中借用这些示例，并通过 @code_lowered 询问 Julia 发生了什么。 :

julia> f() = println("hello world")
julia> g(any_func::Function) = any_func()

#Me: hi, Julia, what happened here?
julia> @code_lowered g(f)

#Julia: hi there, this is the lowered code, anything else you want to know?
CodeInfo(:(begin 
        nothing
        return (any_func)() 
    end))
#Me: it looks like function `g` just returned `(any_func)()` and did nothing else, 
#    is it equvienlent to write `f()`?
julia> @code_typed g(f)

#Julia: yes, the codes looks the same after type inference stage: 
CodeInfo(:(begin 
        $(Expr(:inbounds, false))
        # meta: location REPL[1] f 1
        # meta: location coreio.jl println 5
        SSAValue(0) = (Core.typeassert)(Base.STDOUT, Base.IO)::IO
        # meta: pop location
        # meta: pop location
        $(Expr(:inbounds, :pop))
        return (Base.print)(SSAValue(0), "hello world", $(QuoteNode('\n')))::Void
    end))=>Void

julia> @code_typed f()
CodeInfo(:(begin 
        $(Expr(:inbounds, false))
        # meta: location coreio.jl println 5
        SSAValue(0) = (Core.typeassert)(Base.STDOUT, Base.IO)::IO
        # meta: pop location
        $(Expr(:inbounds, :pop))
        return (Base.print)(SSAValue(0), "hello world", $(QuoteNode('\n')))::Void
    end))=>Void

Does this mean the input function is run before the calling function?

在这种特殊情况下，很难回答，调用函数 g在编译时被编译器优化，所以没有 g在运行时。 :P 让我们添加一些额外的内容到 g :

julia> g(any_func::Function) = (println("I don't wanna be optimized out!"); any_func())
g (generic function with 1 method)
julia> @code_lowered g(f)
CodeInfo(:(begin 
        nothing
        (Main.println)("I don't wanna be optimized out!")
        return (any_func)()
    end))
julia> @code_typed g(f)
CodeInfo(:(begin 
        $(Expr(:inbounds, false))
        # meta: location coreio.jl println 5
        SSAValue(0) = (Core.typeassert)(Base.STDOUT, Base.IO)::IO
        # meta: pop location
        $(Expr(:inbounds, :pop))
        (Base.print)(SSAValue(0), "I don't wanna be optimized out!", $(QuoteNode('\n')))::Void
        $(Expr(:inbounds, false))
        # meta: location REPL[2] f 1
        # meta: location coreio.jl println 5
        SSAValue(1) = (Core.typeassert)(Base.STDOUT, Base.IO)::IO
        # meta: pop location
        # meta: pop location
        $(Expr(:inbounds, :pop))
        return (Base.print)(SSAValue(1), "hello world", $(QuoteNode('\n')))::Void
    end))=>Void

此示例显示 g 的内容将在 f 之前运行，所以答案是肯定的。遵循相同的模式，很容易查看 Liso 的示例:

julia> f = x->x+1
julia> fun(f::Function, a) = f(a)

# nothing new here, `f(9)` was called before `fun` 
# and the return value was passed to `fun` with a binding name `a`.
julia> @code_lowered fun(f, f(9))  
CodeInfo(:(begin 
        nothing
        return (f)(a)
    end))

# to verify your second question:
julia> foo(f, x) = x
foo (generic function with 1 method)

julia> @code_lowered foo(f, 1)
CodeInfo(:(begin 
        nothing
        return x
    end))

Does the input function only get called if the calling function calls it specifically?

所以是的，上面的例子表明如果 f未被调用函数调用 foo ，会直接优化出来。

与其他语言不同，Julia 不仅仅是一个对用户不透明的魔法黑匣子，有时打开盒子自学成才是高效的。顺便说一句，Julia 中还有两个阶段(@code_llvm，@code_native)，您可能需要转储这些低级代码以进行一些高级调查，请参阅 Stefan 在这篇文章中的精彩回答: What is the difference between @code_native, @code_typed and @code_llvm in Julia?了解更多详情。

更新:

what is different between these functions: g1() = f() , g2(f::Function) = f(). Results for both of them are the same, so What is different between them?

julia> @code_lowered g1()
CodeInfo(:(begin 
        nothing
        return (Main.f)()
    end))

julia> @code_lowered g2(f)
CodeInfo(:(begin 
        nothing
        return (f)()
    end))

降低的代码告诉 g1()总是返回 Main.f() ，这里是 Main.f是 f在 Main模块，但 g2返回 f()在哪里 f是您传递给它的函数。为了清楚起见，我们可以定义 g2作为:

julia> g2(argumentf::Function) = argumentf()
g2 (generic function with 1 method)

julia> @code_lowered g2(f)
CodeInfo(:(begin 
        nothing
        return (argumentf)()
    end))

g2是“将函数作为另一个函数的参数传递”， g1可以认为是 Main.f 的别名.你能理解这个吗？ @红色的