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      1. 为什么 sys.getrefcount() 返回 2?

        Why does sys.getrefcount() return 2?(为什么 sys.getrefcount() 返回 2?)
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                  本文介绍了为什么 sys.getrefcount() 返回 2?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

                  问题描述

                  据我了解,sys.getrefcount() 返回对象的引用数,在以下情况下应该"为 1:

                  As I understand, sys.getrefcount() returns the number of references of an object, which "should" be 1 in the following case:

                  import sys,numpy
                  a = numpy.array([1.2,3.4])
                  print sys.getrefcount(a)
                  

                  然而,结果是2!所以,如果我:

                  However, it turned out to be 2! So, if I:

                  del a
                  

                  numpy.array([1.2,3.4])"对象是否仍然存在(没有垃圾回收)?

                  Will the "numpy.array([1.2,3.4])" object still be there (no garbage collection)?

                  推荐答案

                  当你调用 getrefcount() 时,引用按值复制到函数的参数中,暂时增加对象的引用计数.这就是第二个引用的来源.

                  When you call getrefcount(), the reference is copied by value into the function's argument, temporarily bumping up the object's reference count. This is where the second reference comes from.

                  这在文档中有解释:

                  返回的计数通常比您预期的高 1,因为它包含(临时)引用作为参数getrefcount().

                  The count returned is generally one higher than you might expect, because it includes the (temporary) reference as an argument to getrefcount().

                  关于你的第二个问题:

                  如果我del a",numpy.array([1.2,3.4])"对象是否仍然存在(没有垃圾回收)?

                  If I "del a", will the "numpy.array([1.2,3.4])" object still be there (no garbage collection)?

                  getrefcount() 退出时,数组的引用计数将回到 1,随后的 del a 将释放内存.

                  By the time getrefcount() exits, the array's reference count will to back to 1, and a subsequent del a would release the memory.

                  这篇关于为什么 sys.getrefcount() 返回 2?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跟版网!

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