在C ++中i ++和++ i之间有性能差异吗？ [英] Is there a performance difference between i++ and ++i in C++?

问题描述

我们有这个问题 i ++ 和 ++ i 之间有性能差异， code> in C ？

We have the question is there a performance difference between i++ and ++i in C?

C ++的答案是什么？

What's the answer for C++?

推荐答案

[执行摘要：如果您没有特定的理由使用<$ c，请使用 ++ i $ c> i ++ 。]

[Executive Summary: Use ++i if you don't have a specific reason to use i++.]

对于C ++，答案稍微复杂一些。

For C++, the answer is a bit more complicated.

如果 i 是一个简单类型（不是C ++类的实例），然后对C给出的答案（没有没有性能差异）保持，因为编译器正在生成代码。

If i is a simple type (not an instance of a C++ class), then the answer given for C ("No there is no performance difference") holds, since the compiler is generating the code.

如果 i 是C ++类的实例，则 i ++ 和 ++ i 正在调用运算符++ 函数之一。以下是这些函数的标准对：

However, if i is an instance of a C++ class, then i++ and ++i are making calls to one of the operator++ functions. Here's a standard pair of these functions:

Foo& Foo::operator++()   // called for ++i
{
    this->data += 1;
    return *this;
}

Foo Foo::operator++(int ignored_dummy_value)   // called for i++
{
    Foo tmp(*this);   // variable "tmp" cannot be optimized away by the compiler
    ++(*this);
    return tmp;
}

由于编译器不是生成代码， $ c> operator ++ 函数，没有办法优化 tmp 变量及其关联的拷贝构造函数。如果复制构造函数很昂贵，那么这会对性能产生重大影响。

Since the compiler isn't generating code, but just calling an operator++ function, there is no way to optimize away the tmp variable and its associated copy constructor. If the copy constructor is expensive, then this can have a significant performance impact.

（感谢 Paul 查询C和C ++之间的区别。）

(Thanks to Paul for inquiring about the difference between C and C++.)

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