1: var n3 = 52.3685;
2: document.writeln(n3 >> 0);// 52
3: 可以去掉小數。
如上代碼,就是這么簡單
右移位操作導致小數部分丟失,為何會這樣呢?左移位可以嗎?移位操作是否都有如此功能呢?
帶着疑問又寫了一段代碼用來測試以上想法,繼續上代碼
1: {
2: n = 52.123456;
3: //alert(typeof n);
4: alert(n);
5: }
6: //有符號右移
7: {
8: n = 52.123456;
9: var n2 = n >> 0;
10: //alert(typeof n2);
11: alert(n2);
12: }
13: //無符號右移
14: {
15: n = 52.123456;
16: var n3 = n >>> 0;
17: //alert(typeof n3);
18: alert(n3);
19: }
20: //左移0位
21: {
22: n = 52.123456;
23: var n4 = n << 0;
24: //alert(typeof n4);
25: alert(n4);
26: }
27: //按位或or
28: {
29: n = 52.123456;
30: var n5 = n | 0;
31: //alert(typeof n5);
32: alert(n5);
33: }
34: //按位異或xor
35: {
36: n = 52.123456;
37: var n6 = n ^ 0;
38: //alert(typeof n6);
39: alert(n6);
40: }
那,這里不賣關子,直接給出測試結果來:以上五種方法均可以去掉小數點;然而為什么會這樣呢?
翻翻EAMCScript規范吧,或許里邊會有答案,見http://bclary.com/2004/11/07/#a-9.5
先來看看位操作都做了什么,下邊是位操作的實現步驟,重點在第五,第六步
11.7.1 The Left Shift Operator (<< )
Performs a bitwise left shift operation on the left operand by the amount specified by the right operand.
The production ShiftExpression : ShiftExpression << AdditiveExpression is evaluated as follows:
1. Evaluate ShiftExpression.
2.Call GetValue(Result(1)).
3.Evaluate AdditiveExpression.
4. Call GetValue(Result(3)).
5.Call ToInt32(Result(2)).
6.Call ToUint32(Result(4)).
7.Mask out all but the least significant 5 bits of Result(6), that is, compute Result(6) & 0x1F.
8.Left shift Result(5) by Result(7) bits. The result is a signed 32 bit integer.
9.Return Result(8).|
再來看看那鍋ToInt32干了什么,重點在第三步
9.5 ToInt32: (Signed 32 Bit Integer)
The operator ToInt32 converts its argument to one of 232 integer values in the range -231 through 231-1, inclusive. This operator functions as follows:
1. Call ToNumber on the input argument.
2. If Result(1) is NaN, +0, -0, +∞, or -∞, return +0.
3. Compute sign(Result(1)) * floor(abs(Result(1))).
4. Compute Result(3) modulo 232 ; that is, a finite integer value k of Number type with positive sign and less than 232 in magnitude such the mathematical difference of Result(3) and k is mathematically an integer multiple of 232 .
5. If Result(4) is greater than or equal to 231 , return Result(4)-232 , otherwise return Result(4).
最后來看那個floor是什么意思,這里重點看第三步的后半拉,就是那個floor是干什么滴
floor(x) = x-(x modulo 1)
看見沒,就在這一步把小數干掉了
Floor(x) 等於x減去x模上1
即
N= 52.123456 – 52.123456%1
=52.123456-0.1234559999999
=52
搜代斯吶,春節快樂~