重新定义python内置函数 [英] Redefining python built-in function

问题描述

我正在编写一个 python 程序，作者编写了一个看起来像这样的函数

I'm working on a python program and the author has written a function that looks like this

def blah():
    str = "asdf asdf asdf"
    doStuff(str)

这似乎有效，尽管 str 是一个内置函数，不应用作变量.

This seems to work, even though str is a built in function and shouldn't be used as a variable.

这里究竟发生了什么?我的猜测是 str 将不再可用作函数，而只能在他编写的 blah() 函数的范围内使用.那是对的吗?这不会全局重新定义 str ，对吗?

What is actually happening here? My guess is str will no longer be usable as a function, but only in the scope of the blah() function he's written. Is that correct? This won't redefine str globally, right?

推荐答案

在内部，函数的局部变量表将包含 str 的条目，该条目将是该函数的局部变量.您仍然可以通过在 Py3 中执行 builtins.str 和在 Py2 中执行 __builtin__.str 来访问函数中的内置类.函数外的任何代码都看不到函数的任何局部变量，因此内置类可以安全地在其他地方使用.

Internally, the function's local variable table will contain an entry for str, which will be local to that function. You can still access the builtin class within the function by doing builtins.str in Py3 and __builtin__.str in Py2. Any code outside the function will not see any of the function's local variables, so the builtin class will be safe to use elsewhere.

这里还有另一个警告/角落案例，在这个问题中进行了描述.本地表条目是在编译时创建的，而不是在运行时创建的，因此即使在分配 "asdf asdf asdf"str 的全局定义> 给它:

There is another caveat/corner case here, which is described in this question. The local table entry is created at compile-time, not at runtime, so you could not use the global definition of str in the function even before you assign "asdf asdf asdf" to it:

def blah():
    x = str(12)
    str = "asdf asdf asdf"
    doStuff(str)

将失败并返回 UnboundLocalError.

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