C++ – does `const auto` have any meaning

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

The compiler is allowed to make one implicit conversion to resolve the parameters to a function. What this means is that the compiler can use constructors callable with a single parameter to convert from one type to another in order to get the right type for a parameter.

Here's an example class with a constructor that can be used for implicit conversions:

class Foo
{
public:
  // single parameter constructor, can be used as an implicit conversion
  Foo (int foo) : m_foo (foo) 
  {
  }

  int GetFoo () { return m_foo; }

private:
  int m_foo;
};

Here's a simple function that takes a Foo object:

void DoBar (Foo foo)
{
  int i = foo.GetFoo ();
}

and here's where the DoBar function is called:

int main ()
{
  DoBar (42);
}

The argument is not a Foo object, but an int. However, there exists a constructor for Foo that takes an int so this constructor can be used to convert the parameter to the correct type.

The compiler is allowed to do this once for each parameter.

Prefixing the explicit keyword to the constructor prevents the compiler from using that constructor for implicit conversions. Adding it to the above class will create a compiler error at the function call DoBar (42). It is now necessary to call for conversion explicitly with DoBar (Foo (42))

The reason you might want to do this is to avoid accidental construction that can hide bugs.
Contrived example:

• You have a MyString class with a constructor that constructs a string of the given size. You have a function print(const MyString&) (as well as an overload print (char *string)), and you call print(3) (when you actually intended to call print("3")). You expect it to print "3", but it prints an empty string of length 3 instead.