假设Java是实现语言，那么a、b和c都是对初始值为null的Complex类型的引用。还假设Complex是不可变的，就像前面提到的BigInteger和类似的不可变BigDecimal一样，我想你是指下面的意思，因为你将引用赋值给从添加b和c返回的Complex，而不是将这个引用与a进行比较。

不是: 复合物a, b, c;A = b + c; 简单得多: 复合物a, b, c;A = b.add(c);

Saying that operator overloading leads to logical errors of type that operator does not match the operation logic, it's like saying nothing. The same type of error will occur if function name is inappropriate for operation logic - so what's the solution: drop the ability of function usage!? This is a comical answer - "Inappropriate for operation logic", every parameter name, every class, function or whatever can be logicly inappropriate. I think that this option should be available in respectable programing language, and those that think that it's unsafe - hey no bothy says you have to use it. Lets take the C#. They drooped the pointers but hey - there is 'unsafe code' statement - program as you like on your own risk.

我认为这可能是一个有意识的设计选择，迫使开发人员创建的函数的名称清楚地表达了他们的意图。在c++中，开发人员会用与给定操作符的普遍接受性质无关的功能重载操作符，这使得不查看操作符的定义几乎不可能确定一段代码的功能。

Java设计人员认为操作符重载带来的麻烦大于它的价值。就这么简单。

在一种每个对象变量实际上都是引用的语言中，运算符重载具有相当不合逻辑的额外危险——至少对c++程序员是这样。比较c#的==操作符重载和Object的情况。等号和对象。ReferenceEquals(或其他名称)。

从技术上讲，每一种编程语言都有运算符重载，可以处理不同类型的数字，例如整数和实数。解释:术语重载意味着一个函数有多个实现。在大多数编程语言中，为运算符+提供了不同的实现，一个用于整数，一个用于实数，这被称为运算符重载。

Now, many people find it strange that Java has operator overloading for the operator + for adding strings together, and from a mathematical standpoint this would be strange indeed, but seen from a programming language's developer's standpoint, there is nothing wrong with adding builtin operator overloading for the operator + for other classes e.g. String. However, most people agree that once you add builtin overloading for + for String, then it is generally a good idea to provide this functionality for the developer as well.

我完全不同意操作符重载会混淆代码的谬论，因为这是由开发人员决定的。这是naïve的想法，老实说，它已经过时了。

+1用于在Java 8中添加操作符重载。

看看Boost吧。单位:链接文字

它通过操作符重载提供零开销的维数分析。这一点还能清楚多少呢?

quantity<force>     F = 2.0*newton;
quantity<length>    dx = 2.0*meter;
quantity<energy>    E = F * dx;
std::cout << "Energy = " << E << endl;

实际上会输出“Energy = 4 J”，这是正确的。