传递参考到输出位置vs使用返回 [英] Pass reference to output location vs using return

问题描述

当调用一个提供简单数据类型的函数时，如果它填充了一个内存位置（通过指针传递）或者返回了简单数据，那么这样做会更好吗？

Which is better for performance when calling a function that provides a simple datatype -- having it fill in a memory location (passed by pointer) or having it return the simple data?

我过度简化了返回静态值5的示例，但假设确定返回值的查找/功能在现实生活中是动态的...

I've oversimplified the example returning a static value of 5 here, but assume the lookup/functionality that determines the return value would be dynamic in real life...

传统逻辑会告诉我第一种方法更快，因为我们通过引用操作，而不是像第二种方法那样返回一个副本...但是，我想要别人的意见。

Conventional logic would tell me the first approach is quicker since we are operating by reference instead of having to return a copy as in the 2nd approach... But, I'd like others' opinions.

感谢


void func（int * a）{
* a = 5;
}


或...


int func（）{
return 5;
}


int func() { return 5; }

推荐答案

一般来说，函数（也就是返回一个逻辑值），那么最好使用 int func（）。即使返回值是一个复杂的C ++对象，也有一个常见的优化，名为返回值优化，可避免不必要的对象复制，这两种形式在运行时性能上大致相等。

In general, if your function acts like a function (that is, returning a single logical value), then it's probably best to use int func(). Even if the return value is a complex C++ object, there's a common optimisation called Return Value Optimisation that avoids unnecessary object copying and makes the two forms roughly equivalent in runtime performance.

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