如何从C ++函数返回const Float **

I have a class which hold an array "float ** table". Now I want to have member function to return it, but don't want it to be modified outside of the class. So I did this:

我有一个类,其中包含一个数组“float ** table”。现在我希望有成员函数来返回它,但不希望它在类之外被修改。所以我这样做了:

class sometable 
{
  public:
   ...
   void updateTable(......);
   float **getTable() const {return table;}
  private:
    ...
    float **table;
}

This compiles OK when I call getTable with a constant object. Now I tried to make it safer by declaring getTable as "const float **getTable()". I got the following compilation error:

当我使用常量对象调用getTable时,这会编译好。现在我试图通过将getTable声明为“const float ** getTable()”来使其更安全。我收到以下编译错误:

Error:
  Cannot return float**const from a function that should return const float**.

Why? How can I avoid table to be modified out side of the class?

为什么?如何避免将表修改为类的一部分?

4 个解决方案

#1

Declare your method like this:

像这样声明你的方法:

float const* const* getTable() const {return table;}

const float* const* getTable() const {return table;}

if you prefer.

如果你更喜欢。

#2

You can't assign a float** to a float const** because it would allows to modify a const object:

你不能将float **赋给float const **,因为它允许修改const对象:

float const pi = 3.141592693;
float* ptr;
float const** p = &ptr; // example of assigning a float** to a float const**, you can't do that
*p = &pi;  // in fact assigning &pi to ptr
*ptr = 3;  // PI Indiana Bill?

C and C++ rules differ about what is allowed.

C和C ++规则在允许的内容方面有所不同。

C++ rule is that when you add a const before a star, you have to add a const before each following one.

C ++规则是当你在星形之前添加一个const时,你必须在每个后面添加一个const。
C rule is that you can only add a const before the last star.

C规则是你只能在最后一个星之前添加一个const。

In both languages, you can remove a const only before the last star.

在这两种语言中,您只能在最后一颗星之前删除一个const。

#3

You could declare your method as

您可以将方法声明为

const float * const * const getTable() const {return table;}

but even this (the outermost const - next to the function name) would not prevent the client to try to delete it. You could return reference instead, but the best would be to use an std::vector for table and return const ref to it - unless using a C style array is a must

但即便如此(最外面的const - 函数名旁边)也不会阻止客户端尝试删除它。您可以返回引用,但最好的方法是使用std :: vector作为表并将const ref返回给它 - 除非使用C样式数组是必须的

#4

Though you can clearly type the syntax just like that, I find it much more readable to define some typedefs for multiple-dimension arrays.

虽然您可以清楚地键入语法,但我发现为多维数组定义一些typedef更具可读性。

struct M {
    typedef double* t_array;
    typedef const double t_carray;
    typedef t_array* t_matrix;
    typedef const t_carray* t_cmatrix;

    t_matrix values_;

    t_cmatrix values() const { return values_; }
    t_matrix  values()       { return values_; }
};

#1