对于原始类型，+运算符是否重载？ [英] Is + operator overloaded for primitive types?

问题描述

已经重载的运算符<<，>>，=等多次使用。

我想到的一个例子是当我们添加字符串说：

string name = string（munish）+kumar

+运算符在字符串类中重载。

我们添加像1 + 2的数字（看起来不像一个重载的操作符调用）

我只是想知道它是如何发生的编译器做一个二进制加法。 / p>

我不需要担心它，虽然如果编译器这样做，只是一个好奇心为我。

operator + 这实际上是一个奇怪的名称的函数。基本类型的添加由CPU指令执行，例如：

  addl％edx，（％eax）//这将添加两在edx和eax加载的整数值
  

您实现 operator + 用于用户定义的类型，编译器会生成大量CPU指令来执行您在 operator + 中编写的任务。

Already overloaded operators <<,>>,= etc are used many times.

An example that i was thinking of is when we add strings say :

string name = string("munish") + "kumar";

the + operator is overloaded in the string class.

but when we add numbers like 1 + 2( doesn't seem like an overloaded operator call)

I was just wondering how does it happens does the compiler does a binary additon.

I don't need to worry about it much though if the compiler does it so, just a matter of curiosity for me.

解决方案

Primitive types don't implement operator+ which is actually a function with a weird name. Addition for Primitive type is carried out by CPU instruction such as :

addl %edx,(%eax) //this adds two integral values loaded at edx and eax

You implement operator+ for user-defined types, and compiler generates lots of CPU instructions to carry out the task which you write in operator+.

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