I once took a C++ course with someone on the C++ standards committee who implied that Java made the right decision to avoid having unsigned integers because (1) most programs that use unsigned integers can do just as well with signed integers and this is more natural in terms of how people think, and (2) using unsigned integers results in lots easy to create but difficult to debug issues such as integer arithmetic overflow and losing significant bits when converting between signed and unsigned types. If you mistakenly subtract 1 from 0 using signed integers it often more quickly causes your program to crash and makes it easier to find the bug than if it wraps around to 2^32 - 1, and compilers and static analysis tools and runtime checks have to assume you know what you're doing since you chose to use unsigned arithmetic. Also, negative numbers like -1 can often represent something useful, like a field being ignored/defaulted/unset while if you were using unsigned you'd have to reserve a special value like 2^32 - 1 or something similar.

Long ago, when memory was limited and processors did not automatically operate on 64 bits at once, every bit counted a lot more, so having signed vs unsigned bytes or shorts actually mattered a lot more often and was obviously the right design decision. Today just using a signed int is more than sufficient in almost all regular programming cases, and if your program really needs to use values bigger than 2^31 - 1, you often just want a long anyway. Once you're into the territory of using longs, it's even harder to come up with a reason why you really can't get by with 2^63 - 1 positive integers. Whenever we go to 128 bit processors it'll be even less of an issue.

http://skeletoncoder.blogspot.com/2006/09/java-tutorials-why-no-unsigned.html

这个家伙说，因为C标准定义了包含无符号整型和有符号整型的操作被视为无符号整型。这可能导致负符号整数滚动到一个大的无符号整数，可能会导致错误。

你的问题是“为什么Java不支持无符号整数”?

我对你的问题的回答是，Java希望它所有的基本类型:byte, char, short, int和long应该分别被视为字节，word, dword和qword，就像在汇编中一样，Java操作符对所有的基本类型都是有符号的操作，除了char，但只有在char上它们只能是无符号的16位。

因此静态方法对于32位和64位也是无符号操作。

您需要final类，它的静态方法可以用于无符号操作。

你可以创建这个最终类，给它起任何你想要的名字，并实现它的静态方法。

如果你不知道如何实现静态方法，那么这个链接可以帮助你。

在我看来，如果Java既不支持无符号类型，也不支持操作符重载，那么它一点也不类似于c++，所以我认为Java应该被视为与c++和C完全不同的语言。

顺便说一下，这两种语言的名称也完全不同。

所以我不建议在Java中输入类似C的代码也不建议输入类似c++的代码，因为在Java中，你将无法在c++中做你想做的事情，也就是说，代码将不会继续像c++那样，对我来说，这样的代码很糟糕，在中间改变风格。

我建议也为有符号操作编写和使用静态方法，这样就不会在代码中看到有符号和无符号操作混合使用操作符和静态方法，除非在代码中只需要有符号操作，而且只使用操作符是可以的。

此外，我建议避免使用short、int和long基元类型，分别使用word、dword和qword，并且您将调用无符号操作和/或有符号操作的静态方法，而不是使用操作符。

如果您打算只执行有符号的操作，并且只在代码中使用操作符，那么使用这些原语类型short、int和long是可以的。

实际上word, dword和qword在语言中不存在，但你可以为它们创建新类，并且它们的实现应该非常容易:

The class word holds the primitive type short only, the class dword holds the primitive type int only and the class qword holds the primitive type long only. Now all the unsigned and the signed methods as static or not as your choice, you can implement in each class, i.e. all the 16 bit operations both unsigned and signed by giving meaning names on the word class, all the 32 bit operations both unsigned and signed by giving meaning names on the dword class and all the 64 bit operations both unsigned and signed by giving meaning names on the qword class.

如果你不喜欢给每个方法取太多不同的名字，你可以在Java中使用重载，很高兴看到Java没有删除它!

如果您想要8位有符号操作的方法而不是操作符，或者想要8位无符号操作的方法而根本没有操作符，那么您可以创建Byte类(注意，第一个字母'B'是大写的，所以这不是基本类型Byte)并在该类中实现方法。

关于按值传递和按引用传递:

If I am not wrong, like in C#, primitive objects are passed by value naturally, but class objects are passed by reference naturally, so that means that objects of type Byte, word, dword and qword will be passed by reference and not by value by default. I wish Java had struct objects as C# has, so all Byte, word, dword and qword could be implemented to be struct instead of class, so by default they were passed by value and not by reference by default, like any struct object in C#, like the primitive types, are passed by value and not by reference by default, but because that Java is worse than C# and we have to deal with that, then there is only classes and interfaces, that are passed by reference and not by value by default. So if you want to pass Byte, word, dword and qword objects by value and not by reference, like any other class object in Java and also in C#, you will have to simply use the copy constructor and that's it.

这是我能想到的唯一解决办法。我只是希望我可以将原始类型类型定义为word, dword和qword，但Java既不支持typedef，也不支持using，不像c#支持using，这相当于C的typedef。

输出:

对于相同的比特序列，可以以多种方式打印它们:二进制、十进制(就像C printf中%u的含义)、八进制(就像C printf中%o的含义)、十六进制(就像C printf中%x的含义)和整数(就像C printf中%d的含义)。

请注意，C printf不知道作为参数传递给函数的变量的类型，因此printf只知道传递给函数第一个形参的char*对象中的每个变量的类型。

所以在每一个类:Byte, word, dword和qword中，你可以实现print方法并获得printf的功能，即使类的基本类型是有符号的，你仍然可以通过遵循一些涉及逻辑和移位操作的算法将其作为无符号打印到输出中。

不幸的是，我给你的链接没有显示如何实现这些打印方法，但我相信你可以谷歌的算法，你需要实现这些打印方法。

这就是我所能回答你的问题并建议你的。

一旦有符号整型和无符号整型混合在表达式中，事情就开始变得混乱，你可能会丢失信息。将Java限制为有符号int型只能真正解决问题。我很高兴我不必担心整个有符号/无符号的问题，尽管我有时会错过字节中的第8位。

这是一个古老的问题，pat确实简单地提到了char，我只是想我应该为其他人扩展这个问题，他们将在未来的道路上看到这个问题。让我们仔细看看Java的基本类型:

字节- 8位有符号整数

短16位有符号整数

Int - 32位有符号整数

长64位有符号整数

Char - 16位字符(无符号整数)

虽然char不支持无符号算术，但它本质上可以被视为无符号整数。您必须显式地将算术运算转换回char类型，但它确实提供了一种指定无符号数字的方法。

char a = 0;
char b = 6;
a += 1;
a = (char) (a * b);
a = (char) (a + b);
a = (char) (a - 16);
b = (char) (b % 3);
b = (char) (b / a);
//a = -1; // Generates complier error, must be cast to char
System.out.println(a); // Prints ? 
System.out.println((int) a); // Prints 65532
System.out.println((short) a); // Prints -4
short c = -4;
System.out.println((int) c); // Prints -4, notice the difference with char
a *= 2;
a -= 6;
a /= 3;
a %= 7;
a++;
a--;

是的，没有对无符号整数的直接支持(显然，如果有直接支持，我就不必将大部分操作转换回char类型)。但是，肯定存在无符号基元数据类型。我也希望看到一个无符号字节，但我猜加倍内存成本，而不是使用char是一个可行的选择。

Edit

JDK8为Long和Integer提供了新的api，在将Long和int值作为无符号值处理时提供了辅助方法。

compareUnsigned divideUnsigned parseUnsignedInt parseUnsignedLong remainderUnsigned toUnsignedLong toUnsignedString

此外，Guava提供了许多帮助器方法来处理整数类型，这有助于弥补由于缺乏对无符号整数的本机支持而留下的空白。

这是对高斯林和其他人的采访，关于简单:

Gosling: For me as a language designer, which I don't really count myself as these days, what "simple" really ended up meaning was could I expect J. Random Developer to hold the spec in his head. That definition says that, for instance, Java isn't -- and in fact a lot of these languages end up with a lot of corner cases, things that nobody really understands. Quiz any C developer about unsigned, and pretty soon you discover that almost no C developers actually understand what goes on with unsigned, what unsigned arithmetic is. Things like that made C complex. The language part of Java is, I think, pretty simple. The libraries you have to look up.