为什么在某些情况下JavaScript中按0进行按位移位会产生奇怪的结果 [英] Why bitwise shift with 0 in JavaScript yields weird results in some cases

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:

所有按位运算符的操作数均以二进制补码格式转换为带符号的32位整数.

The operands of all bitwise operators are converted to signed 32-bit integers in two's complement format.

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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