在比较Python中的数值时进行隐式转换

Question

在处理一个问题时,我偶然发现了一些我自己无法解决的问题.

我有一个变量: a = pow(2, 1024)

它的类型是long int.如果我尝试将它明确地转换为浮动,就像float(a)我收到的那样OverflowError.这个数字太大了,不适合64位浮点数,所以这是可以理解的.

然后我尝试使用隐式转换,通过浮点数将其拖放:

b = a * 11.0

再次,OverflowError发生时,这是很好的,因为根据python文档,从长整型的隐式转换到浮动发生.结果就像以前一样.

最后,我尝试比较:

a > 11.0 回报True.将OverflowError不会发生.这让我很困惑.如果不需要数字以相同的格式,Python比较机制如何工作？按照这个,

Python完全支持混合算术:当二进制算术运算符具有不同数值类型的操作数时,具有"较窄"类型的操作数被扩展为另一个的操作数,其中普通整数比长整数窄,比浮点更窄,比浮点更窄.复杂.混合类型数量之间的比较使用相同的规则.构造函数int(),long(),float()和complex()可用于生成特定类型的数字.

我的问题是,为什么a不在前面提到的比较中被抛弃？

我使用的Python版本是2.7.15.谢谢你提前

Answer 1

从来源:

/* Comparison is pretty much a nightmare.  When comparing float to float,
 * we do it as straightforwardly (and long-windedly) as conceivable, so
 * that, e.g., Python x == y delivers the same result as the platform
 * C x == y when x and/or y is a NaN.
 * When mixing float with an integer type, there's no good *uniform* approach.
 * Converting the double to an integer obviously doesn't work, since we
 * may lose info from fractional bits.  Converting the integer to a double
 * also has two failure modes:  (1) a long int may trigger overflow (too
 * large to fit in the dynamic range of a C double); (2) even a C long may have
 * more bits than fit in a C double (e.g., on a 64-bit box long may have
 * 63 bits of precision, but a C double probably has only 53), and then
 * we can falsely claim equality when low-order integer bits are lost by
 * coercion to double.  So this part is painful too.
 */

因此,考虑了转换的潜在缺陷.