C++ 通过引用传递数组 [英] C++ pass an array by reference

问题描述

这是否允许通过引用传递数组?

is this allowed to pass an array by reference ?

 void foo(double& *bar) 

似乎我的编译器说不.为什么?通过引用传递数组的正确方法是什么?还是变通?我有一个数组参数，我的方法应该修改它，然后我应该检索它.或者，我可以让这个数组成为一个类成员，这很好用，但它对我的代码的其他部分有很多缺点(我想避免).

Seems that my compiler says no. Why? What is the proper way to pass an array by reference? Or a work around? I have an array argument that my method should modify and that I should retrieve afterwards. Alternatively, I could make this array a class member, which works fine, but it has many drawbacks for other part of my code (that I would like to avoid).

谢谢和问候.

推荐答案

数组只能通过引用传递，实际上:

Arrays can only be passed by reference, actually:

void foo(double (&bar)[10])
{
}

这会阻止您执行以下操作:

This prevents you from doing things like:

double arr[20];
foo(arr); // won't compile

为了能够将任意大小的数组传递给foo，将其设为模板并在编译时捕获数组的大小:

To be able to pass an arbitrary size array to foo, make it a template and capture the size of the array at compile time:

template<typename T, size_t N>
void foo(T (&bar)[N])
{
    // use N here
}

你应该认真考虑使用std::vector，或者如果你有一个支持c++11的编译器，std::array.

You should seriously consider using std::vector, or if you have a compiler that supports c++11, std::array.

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