JAVA逐位代码用途 [英] JAVA Bitwise code purpose , &

问题描述

 //以下代码打印出字母aA bB cC dD eE....类UpCase {公共静态void main(String args []){char ch;for(int i = 0; i< 10; i ++){ch =(字符)('a'+ i);System.out.print(ch);ch =(char)((int)ch& 66503);System.out.print(ch +")}}} 

仍在学习Java，但努力理解按位运算.两种代码都可以工作，但是我不理解这些代码背后的二进制原因.为什么(int)会回退到ch，而66503的用途是什么，使其能够打印出不同的字母大写字母.

 //以下代码显示一个字节内的位Showbits类{公共静态void main(String args []){诠释字节val;val = 123;for(t = 128; t> 0; t = t/2){if((val& t)！= 0)System.out.print("1");否则System.out.print("0");}}}//输出为0 1 1 1 1 0 1 1 

对于此代码的输出，实现它​​的步骤细分是什么?如果123是01111011和128以及64和32是10000000不应输出为00000000?和将0变成0?真的很困惑.

解决方案

第二段代码(显示位):

代码实际上是将十进制转换为二进制.该算法使用了一些魔术，主要是 AND(&)运算符.

考虑数字 123 = 01111011 和 128 = 10000000 .当我们将它们与在一起时，我们得到 0 或非零数字，这取决于 128 中的 1 是否与 1 或 0 .

  10000000&01111011----------00000000 

在这种情况下，答案是 0 ，我们的第一位为0.往前看，我们将 64 = 01000000 ，然后与 123 进行相加.注意 1 向右移动.

  01000000&01111011----------01000000 

这次与123进行与运算将产生一个非零数字，第二位为1.重复此过程.

第一段代码(大写):

此处65503是否定值，.

  32 = 0000 0000 0010 0000〜32 = 1111 1111 1101 1111 

从本质上讲，我们通过对32取反，然后通过与运算从小写字母中减去32.众所周知，从小写ASCII值字符中减去32会将其转换为大写.

// following code prints out Letters aA bB cC dD eE ....

class UpCase {
public static void main(String args[]) {
 char ch;

 for(int i = 0; i < 10; i++) {
  ch = (char)('a' + i);
  System.out.print(ch);

  ch = (char)((int) ch & 66503);

  System.out.print(ch + " ")
  }
 }
}

Still learning Java but struggling to understand bitwise operations. Both codes work but I don't understand the binary reasons behind these codes. Why is (int) casted back to ch and what is 66503 used for that enables it to print out different letter casings.

//following code displays bits within a byte

class Showbits {
  public static void main(String args[]) {
  int t;
  byte val;

  val = 123;
  for(t = 128; t > 0; t = t/2) {
   if((val & t) != 0) 
    System.out.print("1 ");
    else System.out.print("0 ");
   }

  }
 }
 //output is 0 1 1 1 1 0 1 1

For this code's output what's the step breakdown to achieve it ? If 123 is 01111011 and 128 as well as 64 and 32 is 10000000 shouldnt the output be 00000000 ? As & turns anything with 0 into a 0 ? Really confused.

解决方案

Second piece of code(Showbits):

The code is actually converting decimal to binary. The algorithm uses some bit magic, mainly the AND(&) operator.

Consider the number 123 = 01111011 and 128 = 10000000. When we AND them together, we get 0 or a non-zero number depending whether the 1 in 128 is AND-ed with a 1 or a 0.

  10000000
& 01111011
----------
  00000000

In this case, the answer is a 0 and we have the first bit as 0. Moving forward, we take 64 = 01000000 and, AND it with 123. Notice the shift of the 1 rightwards.

  01000000
& 01111011
----------
  01000000

AND-ing with 123 produces a non-zero number this time, and the second bit is 1. This procedure is repeated.

First piece of code(UpCase):

Here 65503 is the negation of 32.

 32 = 0000 0000 0010 0000
~32 = 1111 1111 1101 1111

Essentially, we subtract a value of 32 from the lowercase letter by AND-ing with the negation of 32. As we know, subtracting 32 from a lowercase ASCII value character converts it to uppercase.

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