问题描述
只是在JavaScript中进行了异常的按位操作，在某些情况下，我得到了一些奇怪的结果:
Just played around with unusual bitwise operations in JavaScript and I got some weird results in some cases:
常用案例
Usual cases
1 << 0 // returns 1, makes sense
100 << 0 // returns 100, makes sense
100 >> 0 // returns 100, definitely makes sense
但是这些，当移位0位时，全都为零
But these, when shift by 0 bits, all yield zero
9E99 << 0 // returns 0 ..... Why all bits are cleared?
9E99 >> 0 // returns 0 also ..... All bits cleared?
Infinity >> 0 // returns 0
Infinity << 0 // returns 0
-Infinity << 0 // returns 0 .... Can't explain why
-0 << 0 // also yields 0 not -0 itself
-0 >> 0 // also resolved to 0
如果无穷大和按位移位
What if Infinity and bitwise shift
1 << Infinity // returns 1 .. no changes
1024 << Infinity // returns 1024 .. no changes
1024 >> Infinity // returns 1024 .. no changes either
Infinity >> Infinity // 0
Infinity << Infinity // 0
上述情况对我来说没有多大意义.将整数移位零位时，值不变. 但是当您将 Infinity 移0位时，实际上会返回0 .为什么?
Those cases above don't make much sense to me. When shift an integer by zero bits, the value doesn't change. But when you shift Infinity by 0 bits, it actually returns you 0. Why?
我认为将任何数值都移位0位应该不会更改其值，不是吗?
I thought shifting any numeric value by 0 bit should not change its value, doesn't it?
此外，当将一个小的整数值移位无穷大位时，该值完全不变.但是，当您将无穷大移位任何值时，它将被0代替.
Further, when shift a small integer value by infinity bits, the value doesn't change at all. But when you shift Infinity by any values, it is replaced by 0 instead.
我真的很好奇为什么会发生这些现象?是否有任何规范或理论可以解释这些奇怪的行为?
I'm really curious why these phenomena happen? Is there any specifications or theory that explain these odd behaviors?
推荐答案
来自MDN:
JavaScript中的所有数字均为 IEEE754双精度浮点数.在应用按位运算符之前，它们已转换为32位整数.
All numbers in JavaScript are IEEE754 double precision floating point numbers. They're converted to 32 bits integers before being applied bitwise operators.
更准确地说，此过程在 ToInt32中进行了描述(在规范中):
More precisely, this process is described in ToInt32 in the spec:
如您所见，精确描述了Infinities和 -0 的转换.
As you see, the transformation of Infinities and -0 is precisely described.
这是一个二进制截断，这也解释了为什么将 9E99 更改为 0 的原因:如果您查看右边的32位，则最明显>(9E99).toString(2)(将其粘贴到控制台中，它们都是 0 ).
This is a binary truncation, which also explains why 9E99 is changed into 0: This is most obvious if you look at the 32 bits at the right of (9E99).toString(2) (paste it in your console, they're all 0).
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