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Unsigned short in Java

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问题 How can I declare an unsigned short value in Java? 回答1: You can't, really. Java doesn't have any unsigned data types, except char . Admittedly you could use char - it's a 16-bit unsigned type - but that would be horrible in my view, as char is clearly meant to be for text: when code uses char , I expect it to be using it for UTF-16 code units representing text that's interesting to the program, not arbitrary unsigned 16-bit integers with no relationship to text. 回答2: If you really need a

What is short circuiting and how is it used when programming in Java? [duplicate]

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Possible Duplicate: Does java evaluate remaining conditions after boolean result is known Why do we usually use || not | , what is the difference? I missed my class lecture the other day and I was wondering if anyone could give an explanation what short circuiting is and maybe an example of it being used in a simple Java program. Thanks for your help! T.J. Crowder Short-circuiting is where an expression is stopped being evaluated as soon as its outcome is determined. So for instance: if (a == b || c == d || e == f) { // Do something } If a == b is true, then c == d and e == f are never

Primitive type 'short' - casting in Java

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I have a question about the primitive type short in Java. I am using JDK 1.6. If I have the following: short a = 2; short b = 3; short c = a + b; the compiler does not want to compile - it says that it "cannot convert from int to short" and suggests that I make a cast to short , so this: short c = (short) (a + b); really works. But my question is why do I need to cast? The values of a and b are in the range of short - the range of short values is {-32,768, 32767}. I also need to cast when I want to perform the operations -, *, / (I haven't checked for others). If I do the same for primitive

Overflowing Short in java

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问题 I have one question about the short data type in Java. I know that the range for short is between -32768 to 32767. So, if I tried to add two short values that exceed the range, the result ends up being the supposed total minus either the positive range or the negative range times 2, as the following: short a = 30000; a = (short) (a+a); the result is -5536 . So the math is 32768 + 32768 = 65536 , 6000 - 65536 = -5536 . I know what it does, but I don\'t know why it does it this way. Can anybody

2 bytes to short java

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问题 i\'m reading 133 length packet from serialport,last 2 bytes contain CRC values,2 bytes value i\'ve make single(short i think) using java. this what i have done, short high=(-48 & 0x00ff); short low=80; short c=(short) ((high<<8)+low); but i\'m not getting correct result,is it problem because signed valued? how can i solve this problem,plz help me i\'m in trouble 回答1: Remember, you don't have to tie yourself in knots with bit shifting if you're not too familiar with the details. You can use a

Java's L number (long) specification

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问题 It appears that when you type in a number in Java, the compiler automatically reads it as an integer, which is why when you type in (long) 6000000000 (not in integer\'s range) it will complain that 6000000000 is not an integer. To correct this, I had to specify 6000000000L . I just learned about this specification. Are there other number specifications like for short, byte, float, double? It seems like these would be good to have because (I assume) if you could specify the number you\'re

Primitive type 'short' - casting in Java

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问题 I have a question about the primitive type short in Java. I am using JDK 1.6. If I have the following: short a = 2; short b = 3; short c = a + b; the compiler does not want to compile - it says that it \"cannot convert from int to short\" and suggests that I make a cast to short , so this: short c = (short) (a + b); really works. But my question is why do I need to cast? The values of a and b are in the range of short - the range of short values is {-32,768, 32767}. I also need to cast when I

Integer summing blues, short += short problem

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问题 Program in C#: short a, b; a = 10; b = 10; a = a + b; // Error : Cannot implicitly convert type \'int\' to \'short\'. // we can also write this code by using Arithmetic Assignment Operator as given below a += b; // But this is running successfully, why? Console.Write(a); 回答1: There are two questions here. The first is "why is short plus short result in int?" Well, suppose short plus short was short and see what happens: short[] prices = { 10000, 15000, 11000 }; short average = (prices[0] +

Why must a short be converted to an int before arithmetic operations in C and C++?

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问题 From the answers I got from this question, it appears that C++ inherited this requirement for conversion of short into int when performing arithmetic operations from C. May I pick your brains as to why this was introduced in C in the first place? Why not just do these operations as short ? For example ( taken from dyp\'s suggestion in the comments ): short s = 1, t = 2 ; auto x = s + t ; x will have type of int . 回答1: If we look at the Rationale for International Standard—Programming