Java中的C shift运算符的等价物？

发布时间：2020-12-14 17:48:51 所属栏目：Java 来源：网络整理

导读：C移位运算符不循环例如,如果你这样做： // C++int a = 1;cout (a38); 你得到0.但是,在Java中,你实际上会循环并获得64的有效值. 我需要将一些C代码翻译成Java,所以我用什么作为？解决方法 Java language spec说： If the promoted type of the left-hand ope

C移位运算符<不循环例如,如果你这样做：

// C++
int a = 1;
cout << (a<<38);

你得到0.但是,在Java中,你实际上会循环并获得64的有效值.

我需要将一些C代码翻译成Java,所以我用什么作为<<？

解决方法

Java language spec说：

If the promoted type of the left-hand operand is int,only the five
lowest-order bits of the right-hand operand are used as the shift
distance. It is as if the right-hand operand were subjected to a
bitwise logical AND operator & (§15.22.1) with the mask value 0x1f
(0b22222). The shift distance actually used is therefore always in the
range 0 to 31,inclusive.

If the promoted type of the left-hand operand is long,then only the
six lowest-order bits of the right-hand operand are used as the shift
distance. It is as if the right-hand operand were subjected to a
bitwise logical AND operator & (§15.22.1) with the mask value 0x3f
(0b222221). The shift distance actually used is therefore always in
the range 0 to 63,inclusive.

所以,在你的例子中,(int)(((long)a)<< 38)应该工作.

（编辑：李大同）

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