[英]I add two numbers with the same way but get different values
the following is the code and I am extremely confused with the results I get. 以下是代码,我对获得的结果感到非常困惑。
byte[] bytes = new byte[2];
bytes[0] = (byte)0xE6;
bytes[1] = (byte)0x1B;
int high = (bytes[1] & 0xFF)*256;
int low = bytes[0] & 0xFF;
double j = high + low;
double i = (bytes[1] & 0xFF)*256 + bytes[0] & 0xFF;
System.out.println(bytes[1] & 0xFF);
System.out.println(bytes[0] & 0xFF);
System.out.println((bytes[1] & 0xFF)*256);
System.out.println(i);
System.out.println(j);
Logically, i
and j
are the same but the result I get is quite amazing. 从逻辑上讲, i
和j
是相同的,但是我得到的结果非常惊人。
Results: 结果:
27
230
6912
230.0
7142.0
i
and j
should be the same but it isn't. i
和j
应该相同,但事实并非如此。 I don't know the reason. 我不知道原因 Is there any explanation for this? 有什么解释吗?
They aren't equivalent; 它们不是等效的。 In java, the bitwise &
operator has a lower precedence than the +
operator . 在Java中, 按位&
运算符的优先级低于+
运算符 。 So i
is actually being defined as ((bytes[1] & 0xFF)*256 + bytes[0])& 0xFF
, which evaluates to 230, instead of the intended 7142. Replacing that line with i = (bytes[1] & 0xFF)*256 + (bytes[0] & 0xFF);
因此, i
实际上定义为((bytes[1] & 0xFF)*256 + bytes[0])& 0xFF
,其结果为230,而不是预期的7142。用i = (bytes[1] & 0xFF)*256 + (bytes[0] & 0xFF);
is all you need to do. 这就是您需要做的。
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