= Java中的运算符 [英] = operator in java

问题描述

有人可以向我解释为什么可以这样做:

can somebody explain me why it's possible to do:

String s = "foo";

在没有操作符重载的情况下(在这种情况下为"=)

how is this possible without operator overloading (in that case the "=")

我来自C ++背景，这可以说明...

I'm from a C++ background so that explains...

推荐答案

在这种情况下，没有重载.与C ++不同的Java片段是"的定义-Java编译器将"中的任何内容转换为java.lang.string，因此在您的示例中也是一个简单的赋值.在C ++中，编译器将"转换为char const *，因此需要从char const *转换为std :: string.

In this case there is no overloading. The java piece that differs from C++ is the definition of "" - The java compiler converts anything in "" into a java.lang.string and so is a simple assignment in your example. In C++ the compiler converts "" into a char const * and so needs to have a conversion from char const* to std::string.

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