字符串是否在 Python 中池化? [英] Are strings pooled in Python?

问题描述

Python 是否有一个包含所有字符串的池并且它们(字符串)是单例的吗?

Does Python have a pool of all strings and are they (strings) singletons there?

更准确地说，在下面的代码中，是在内存中创建了一个还是两个字符串?

More precise, in the following code, are one or two strings created in memory?

a = str(num)
b = str(num)

推荐答案

字符串在 Python 中是不可变的，因此实现可以决定是否实习(这是一个经常与 C# 相关的术语，意思是一些字符串存储在池中)字符串与否.

Strings are immutable in Python, so the implementation can decide whether to intern (that's a term often associated with C#, meaning that some strings are stored in a pool) strings or not.

在您的示例中，您正在动态创建字符串.CPython并不总是查看池以检测字符串是否已经存在 - 这也没有意义，因为您首先必须保留内存以创建字符串，然后将其与池内容(对于长字符串效率低下).

In your example, you're dynamically creating strings. CPython does not always look into the pool to detect whether the string is already there - it also doesn't make sense because you first have to reserve memory in order to create the string, and then compare it to the pool content (inefficient for long strings).

但是对于长度为 1 的字符串，CPython 确实会查看池(参见stringobject.c"):

But for strings of length 1, CPython does look into the pool (cf. "stringobject.c"):

static PyStringObject *characters[UCHAR_MAX + 1];

...

PyObject *
PyString_FromStringAndSize(const char *str, Py_ssize_t size)
{

...

    if (size == 1 && str != NULL &&
    (op = characters[*str & UCHAR_MAX]) != NULL)
    {
        #ifdef COUNT_ALLOCS
            one_strings++;
        #endif

        Py_INCREF(op);
        return (PyObject *)op;
    }

...

所以:

a = str(num)
b = str(num)
print a is b # <-- this will print False in most cases (but try str(1) is str(1))

但是当在代码中直接使用 constant 字符串时，CPython 使用相同的字符串实例:

But when using constant strings directly in your code, CPython uses the same string instance:

a = "text"
b = "text"
print a is b # <-- this will print True

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