函数参数：指向对象数组的指针

In my main function I create an array of objects of a certain class "Menu"

在我的main函数中,我创建了一个特定类“Menu”的对象数组

And when I call a function I want to provide a pointer to that array.

当我调用一个函数时,我想提供一个指向该数组的指针。

Menu menu[2];
// Create menu [0], [1]
Function(POINTER_TO_ARRAY);

Question: What is the correct way to write the Function parameters?

问题:编写Function参数的正确方法是什么?

I try:

Function(&menu);

and in Header file:

并在头文件中:

void Function(Menu *menu[]); // not working
error: Cannot convert parameter 1 from Menu(*)[2] to Menu *[]

void Function(Menu * menu); // not working
error: Cannot convert parameter 1 from Menu(*)[2] to Menu *[]

and I can't come up with any other way to do this and I can't find a solution to this particular problem.

我无法想出任何其他方法来做到这一点,我无法找到解决这一特定问题的方法。

Simply, I want to be able to access the Menu array within the function through a pointer. What are the difference in normal pointer to a pointer to an array?

简单地说,我希望能够通过指针访问函数内的Menu数组。指向数组的指针的普通指针有什么区别?

4 个解决方案

#1

Declaration:

void Function(Menu* a_menus); // Arrays decay to pointers.

Invocation:

Function(menu);

However, you would need to inform Function() how many entries are in the array. As this is C++ suggest using std::array or std::vector which have knowledge of their size, beginning and end:

但是,您需要通知Function()数组中有多少条目。因为这是C ++建议使用std :: array或std :: vector,它们知道它们的大小,开头和结尾:

std::vector<Menu> menus;
menus.push_back(Menu("1"));
menus.push_back(Menu("2"));

Function(menus);

void Function(const std::vector<Menu>& a_menus)
{
    std::for_each(a_menus.begin(),
                  a_menus.end(),
                  [](const Menu& a_menu)
                  {
                      // Use a_menu
                  });
}

#2

Either by const or non-const pointer

通过const或非const指针

void Function(Menu const* menu);
void Function(Menu* menu);

...or by const or non-const reference

...或通过const或非const引用

void Function(Menu const (&menu)[2]);
void Function(Menu (&menu)[2]);

which can be generalized to a template so that the array size will be deduced by the compiler:

可以推广到模板,以便编译器推导出数组大小:

template<size_t N> void Function(Menu const (&menu)[N]);
template<size_t N> void Function(Menu (&menu)[N]);

Always call as Function(menu);

始终调用函数(菜单);

#3

Should work if you use

如果你使用,应该工作

 void Function(Menu * menu);

and call using

并使用

Function(menu);

instead of

Function(&menu);

passing the array name causes it to decay to a pointer to the type contained in the array. However, as @hmjd says in his answer you will also need to pass the array size, so his suggestion of using a vector is favourable if this option is open to you.

传递数组名称会导致它衰减为指向数组中包含的类型的指针。但是,正如@hmjd在他的回答中所说,你还需要传递数组大小,所以如果这个选项对你开放,他建议使用向量是有利的。

#4

You can use

您可以使用

Function((void *) whatever_pointer_type_or_array_of_classes);

in your main.

在你的主要。

And in the function:

并在功能:

type Function(void * whatever)
{
    your_type * x =(your_type *) whatever;
    //use x 
    ....
    x->file_open=true;
    ....
}

#1