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

问题描述

这是允许按引用传递数组？

is this allowed to pass an array by reference ?

 void foo(double& *bar) 

看来，我的编译器说，没有。为什么？什么是按引用传递数组的正确方法？或周围的工作？我有我的方法应该修改，我应该事后检索数组参数。或者，我可以让这个数组类的成员，它工作得很好，但它对于我的code的另一部分（即我想避免）。

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).

感谢和问候。

推荐答案

数组只能通过引用传递，实际上是：

Arrays can only be passed by reference, actually:

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

这prevents你做这样的事情：

This prevents you from doing things like:

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

要能够为任意大小的数组传递给富，使它成为一个模板，并捕获在编译时数组的大小：

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
}

您应该认真考虑使用的std ::矢量，或者如果你有一个支持C ++ 11，的std ::阵列编译器。

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

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