问题描述
似乎有理由相信以原子方式运行，因为如果指定的键丢失并且没有提供默认值，则它会引发 KeyError
d.pop(k)
但是，该文档似乎并没有具体说明这一点，至少在本节不具体记录 dict.pop 。
However, the documentation does not appear to specifically address that point, at least not in the section specifically documenting dict.pop.
我正在审查使用了这种模式，这个问题发生在我身上：
This question occurred to me as I was reviewing an answer of mine which used this pattern:
if k in d: del d[k]
当时，我并没有想到在中如果可能存在密钥的潜在条件，而不是在德尔。如果 dict.pop 确实提供了一个原子替代方法，那么我应该注意到在我的答案中。
At the time, I was not thinking of the potential condition that a key may be present during the if, but not at the time of del. If dict.pop does indeed provide an atomic alternative, then I should note that in my answer.
推荐答案
对于默认类型， dict.pop（）是一个C函数调用，这意味着它使用一个字节码评估。这使得该调用成为原子。
For the default type, dict.pop() is a C-function call, which means that it is executed with one bytecode evaluation. This makes that call atomic.
只有当字节码评估循环允许它们时，Python线程才会切换，因此在字节码边界处。一些Python C函数会调用Python代码（认为 __ dunder __ 特殊方法钩子），但是 dict.pop（）方法没有，至少不是默认的 dict 类型。
Python threads switch only when the bytecode evaluation loop lets them, so at bytecode boundaries. Some Python C functions do call back into Python code (think __dunder__ special method hooks), but the dict.pop() method does not, at least not for the default dict type.
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