在Python中使函数变懒

Question

我想为eternity编写一个包装器函数，使其表现得像是在懒洋洋地求值。换句话说，它的功能应该与下面的函数lazyeternity相同。我有一个函数packageN，它接受一个函数并将其打包到lambda中——或者至少我认为是这样。（将lazyeternity封装在lambda中延迟了对etest的调用）如何修改packageN以便eternity = packageN(eternity)能够反映{}的行为？在

    def eternity():
        return eternity()

    # How can I create a wrapper function for eternity...
    # that will make it behave identical to this function?
    def lazyeternity():
        return lambda: lazyeternity()

    def packageN(f):
        return lambda *x: f(*x)

    def etest(x, y):
        return x

    eternity = packageN(eternity)

    # I want these both to return 4.
    # Currently only the first one returns 4,...
    # because the second one evaluates the eternity()...
    # parameter forever. I want to delay the evaluation...
    # of the parameter.
    print etest(4, lazyeternity())
    print etest(4, eternity())

Answer 1

要使eternity都工作，只需删除{}。您想要返回一个函数，而不是函数的求值（无限循环）。在

Answer 2

packageN返回的lazyeternity和{}之间的区别在于，调用lazyeternity返回lambda，而{}是一个lambda，当被调用时，它将永远运行。在

要使packageN返回类似lazyeternity的内容，请将其

def packageN(f):
    return lambda: lambda *x: f(*x)