C++ 中的 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

我们有一个问题 i++ 和 ++i 在 C 中是否有性能差异强>?

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:如果您没有特定原因使用 i++，请使用 ++i>.]

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

对于 C++，答案有点复杂.

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)，那么给出的答案对于 C(不，没有性能差异") 成立，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.

但是，如果 i 是 C++ 类的实例，则 i++ 和 ++i 正在调用 i++i 之一代码>operator++ 函数.这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;
}

由于编译器不生成代码，而只是调用 operator++ 函数，因此无法优化掉 tmp 变量及其关联的复制构造函数.如果复制构造函数的开销很大，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.

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