C++ Initializing static const structure variable

cconstantsinitializationstaticstruct

I'm trying to add a static constant variable to my class, which is an instance of a structure. Since it's static, I must initialize it in class declaration. Trying this code

class Game {
    public:
        static const struct timespec UPDATE_TIMEOUT = { 10 , 10 };

    ...
};

Getting this error:

error: a brace-enclosed initializer is not allowed here before '{'
token

error: invalid in-class initialization of static data member of non-integral type 'const timespec'

How do I initialize it? Thanks!

Best Answer

Initialize it in a separate definition outside the class, inside a source file:

// Header file
class Game {
    public:
        // Declaration:
        static const struct timespec UPDATE_TIMEOUT;
    ...
};

// Source file
const struct timespec Game::UPDATE_TIMEOUT = { 10 , 10 };  // Definition

If you include the definition in a header file, you'll likely get linker errors about multiply defined symbols if that header is included in more than one source file.

Related Solutions

C# – the difference between const and readonly in C#

Apart from the apparent difference of

having to declare the value at the time of a definition for a const VS readonly values can be computed dynamically but need to be assigned before the constructor exits.. after that it is frozen.
const's are implicitly static. You use a ClassName.ConstantName notation to access them.

There is a subtle difference. Consider a class defined in AssemblyA.

public class Const_V_Readonly
{
  public const int I_CONST_VALUE = 2;
  public readonly int I_RO_VALUE;
  public Const_V_Readonly()
  {
     I_RO_VALUE = 3;
  }
}

AssemblyB references AssemblyA and uses these values in code. When this is compiled:

in the case of the const value, it is like a find-replace. The value 2 is 'baked into' the AssemblyB's IL. This means that if tomorrow I update I_CONST_VALUE to 20, AssemblyB would still have 2 till I recompile it.
in the case of the readonly value, it is like a ref to a memory location. The value is not baked into AssemblyB's IL. This means that if the memory location is updated, AssemblyB gets the new value without recompilation. So if I_RO_VALUE is updated to 30, you only need to build AssemblyA and all clients do not need to be recompiled.

So if you are confident that the value of the constant won't change, use a const.

public const int CM_IN_A_METER = 100;

But if you have a constant that may change (e.g. w.r.t. precision).. or when in doubt, use a readonly.

public readonly float PI = 3.14;

Update: Aku needs to get a mention because he pointed this out first. Also I need to plug where I learned this: Effective C# - Bill Wagner

C++ – What are the differences between a pointer variable and a reference variable in C++

A pointer can be re-assigned:

int x = 5;
int y = 6;
int *p;
p = &x;
p = &y;
*p = 10;
assert(x == 5);
assert(y == 10);

A reference cannot be re-bound, and must be bound at initialization:

int x = 5;
int y = 6;
int &q; // error
int &r = x;

A pointer variable has its own identity: a distinct, visible memory address that can be taken with the unary & operator and a certain amount of space that can be measured with the sizeof operator. Using those operators on a reference returns a value corresponding to whatever the reference is bound to; the reference’s own address and size are invisible. Since the reference assumes the identity of the original variable in this way, it is convenient to think of a reference as another name for the same variable.
```
int x = 0;
int &r = x;
int *p = &x;
int *p2 = &r;

assert(p == p2); // &x == &r
assert(&p != &p2);
```

You can have arbitrarily nested pointers to pointers offering extra levels of indirection. References only offer one level of indirection.

int x = 0;
int y = 0;
int *p = &x;
int *q = &y;
int **pp = &p;

**pp = 2;
pp = &q; // *pp is now q
**pp = 4;

assert(y == 4);
assert(x == 2);

A pointer can be assigned nullptr, whereas a reference must be bound to an existing object. If you try hard enough, you can bind a reference to nullptr, but this is undefined and will not behave consistently.

/* the code below is undefined; your compiler may optimise it
 * differently, emit warnings, or outright refuse to compile it */

int &r = *static_cast<int *>(nullptr);

// prints "null" under GCC 10
std::cout
    << (&r != nullptr
        ? "not null" : "null")
    << std::endl;

bool f(int &r) { return &r != nullptr; }

// prints "not null" under GCC 10
std::cout
    << (f(*static_cast<int *>(nullptr))
        ? "not null" : "null")
    << std::endl;

You can, however, have a reference to a pointer whose value is nullptr.

Pointers can iterate over an array; you can use ++ to go to the next item that a pointer is pointing to, and + 4 to go to the 5th element. This is no matter what size the object is that the pointer points to.
A pointer needs to be dereferenced with * to access the memory location it points to, whereas a reference can be used directly. A pointer to a class/struct uses -> to access its members whereas a reference uses a ..
References cannot be put into an array, whereas pointers can be (Mentioned by user @litb)
Const references can be bound to temporaries. Pointers cannot (not without some indirection):
```
const int &x = int(12); // legal C++
int *y = &int(12); // illegal to take the address of a temporary.
```
This makes const & more convenient to use in argument lists and so forth.

Best Answer

Related Solutions

C# – the difference between const and readonly in C#

C++ – What are the differences between a pointer variable and a reference variable in C++

Related Topic