为什么Java不支持无符号整数?

在我看来,这是一个奇怪的遗漏,因为它们允许人们编写不太可能在意外的大输入上产生溢出的代码。

此外,使用无符号整数可以是一种自我文档的形式,因为它们表明无符号整型所要保存的值永远不应该是负数。

最后,在某些情况下,无符号整数对于某些运算(如除法)更有效。

包含这些的缺点是什么?


当前回答

我听说它们将在接近最初的Java发行版时被包含。Oak是Java的前身,在一些规范文档中提到了使用值。不幸的是,这些都没有被引入Java语言。据任何人所知,可能由于时间限制,它们没有得到实施。

其他回答

恕我直言,原因是他们太懒了,没有去实施/纠正这个错误。 暗示C/ c++程序员不理解unsigned,结构,联合,位标志…太荒谬了。

如果你正在和一个基本的/bash/java程序员交谈,即将开始用C语言编程,没有任何真正的语言知识,或者你只是在说你自己的想法。;)

当你每天处理文件或硬件的格式时,你会开始质疑,他们到底在想什么。

一个很好的例子是尝试使用无符号字节作为自旋转循环。 对于那些不理解最后一句话的人,你究竟是如何称自己为程序员的。

DC

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.

字里行间,我认为逻辑是这样的:

通常,Java设计人员希望简化可用的数据类型 对于日常用途,他们认为最常见的需求是有符号的数据类型 为了实现某些算法,有时需要无符号算术,但是要实现这种算法的程序员也应该具备使用有符号数据类型进行无符号算术的知识

总的来说,我认为这是一个合理的决定。我可能会:

使字节无符号,或者至少为这一数据类型提供了有符号/无符号的替代选项,可能使用不同的名称(使它有符号有利于一致性,但什么时候需要有符号字节?) 不再使用“short”(你上次使用16位符号算术是什么时候?)

不过,只要稍加修饰,对32位以内的无符号值进行运算就不会太糟糕,而且大多数人不需要无符号64位除法或比较。

因为无符号类型是纯粹的邪恶。

事实上,在C语言中unsigned int生成unsigned更是邪恶的。

下面是一个让我不止一次头疼的问题的快照:

// We have odd positive number of rays, 
// consecutive ones at angle delta from each other.
assert( rays.size() > 0 && rays.size() % 2 == 1 );

// Get a set of ray at delta angle between them.
for( size_t n = 0; n < rays.size(); ++n )
{
    // Compute the angle between nth ray and the middle one.
    // The index of the middle one is (rays.size() - 1) / 2,
    // the rays are evenly spaced at angle delta, therefore
    // the magnitude of the angle between nth ray and the 
    // middle one is: 
    double angle = delta * fabs( n - (rays.size() - 1) / 2 ); 

    // Do something else ...
}

你注意到这个bug了吗?我承认我是在使用调试器之后才看到它的。

由于n是无符号类型size_t,整个表达式n - (ray .size() - 1) / 2的计算结果为无符号。该表达式旨在表示从中间那条线开始的第n条射线的符号位置:从左边那条线开始的第1条射线的位置为-1,右边那条线的位置为+1,等等。在取abs值并乘以角之后,我将得到第n条射线与中间那条射线之间的夹角。

不幸的是,对我来说,上面的表达式包含了邪恶的unsigned,它的计算结果不是-1,而是2^32-1。随后转换为双密封的bug。

由于滥用无符号算术而导致的一两个错误之后,人们不得不开始考虑获得的额外比特是否值得额外的麻烦。我正在尽可能地避免在算术中使用无符号类型,尽管仍然将它用于非算术操作,如二进制掩码。

我能想到一个不幸的副作用。在java嵌入式数据库中,使用32位id字段可以拥有的id数量是2^31,而不是2^32(~ 20亿,而不是~ 40亿)。