Python:重新定义函数，使其引用自己的自身

If I now want process to take the place of fun for all future calls to fun, I need to do something like# Does not workfun = process但是，这不起作用，因为这会产生一个循环引用问题，因为现在从 fun 的正文中调用了 fun .我发现的一种解决方案是在 process 内引用 fun 的副本，如下所示:This does not work however, as this creates a cyclic reference problem as now fun is called from within the body of fun. One solution I have found is to reference a copy of fun inside of process, like so:# Worksimport copyfun_cp = copy.copy(fun)def process(x): x += 1 return fun_cp(x)fun = process但是这个解决方案使我感到困扰，因为我真的不知道Python如何构造函数的副本.我猜我的问题与使用继承和 super 函数扩展类方法的问题相同，但是在这里我没有类.but this solution bothers me as I don't really know how Python constructs a copy of a function. I guess my problem is identical to that of extending a class method using inheritance and the super function, but here I have no class.如何正确执行此操作?我认为这是一项足够常见的任务，应该或多或少有一些惯用的解决方案，但是我没有运气找到它.How can I do this properly? I would think that this is a common enough task that some more or less idiomatic solution should exist, but I have had no luck finding it.推荐答案 Python不是 构造函数的副本. copy.copy(fun)仅返回 fun ;区别在于，您将其保存到 fun_cp 变量中，该变量与您将 process 保存到的变量不同，因此当它保存在 fun_cp 中时，进程尝试寻找它.Python is not constructing a copy of your function. copy.copy(fun) just returns fun; the difference is that you saved that to the fun_cp variable, a different variable from the one you saved process to, so it's still in fun_cp when process tries to look for it.我将执行与您所做的类似的操作，将原始函数保存到其他变量中，而无需使用复制":I'd do something similar to what you did, saving the original function to a different variable, just without the "copy":original_fun = fundef fun(x): x += 1 return original_fun(x)如果要将相同的包装应用于多个函数，则定义装饰器并执行 fun = decorate(fun)更具可重用性，但一次性而言，它比必要的工作多，并且额外的缩进级别.If you want to apply the same wrapping to multiple functions, defining a decorator and doing fun = decorate(fun) is more reusable, but for a one-off, it's more work than necessary and an extra level of indentation. 这篇关于Python:重新定义函数，使其引用自己的自身的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！