P>Собственно вот такой вопрос и есть..

const long val = 10;
...
(long)val = 15;

const long val = 10;
...
const_cast<long>(val) = 15;

MN>>const long val = 10;
MN>>...
MN>>const_cast<long>(val) = 15;
MN>>

X x;
X const& y = x;
X& z = const_cast<X &>(y); // well-formed, well-defined

P>Собственно вот такой вопрос и есть..

5.2.9 Static cast [expr.static.cast]
1 The result of the expression static_cast<T>(v) is the result of converting the expression v to type T.
If T is a reference type, the result is an lvalue; otherwise, the result is an rvalue. Types shall not be defined in a static_cast. The static_cast operator shall not cast away constness (5.2.11).

struct St {int v;St(int);};

void f()
{ typedef int T;
  //typedef St T;

  T a=1;
  static_cast<T>(a)=__LINE__; //expression must be a modifiable lvalue
  (T)a=__LINE__; //expression must be a modifiable lvalue

  T(a)=__LINE__; // "a" has already been declared in the current scope
  {
    T(c)=__LINE__;
    T(b)=__LINE__;
    c=b;
    c=T(b);
  } 
}

void f()
{ int a;
  int(a);//error: "a" has already been declared in the current scope
  int((int)a);// warning: expression has no effect
  int(0);// warning
  int;//warning: ignored on left of 'int' when no variable is declared
}

P>>  int a;   // обьявление переменной a..
P>>  int(a);  // еще обно обьявление переменной а

P>  T a=1;
P>  T(a)=__LINE__; // "a" has already been declared in the current scope[/b]

1, int( a );

struct Struct
{  Struct(int,int)
    {
    }
  Struct(int)
    {
    }
  Struct()
    {
    }
};


int main(int , char **)
{   int aa;
    {  Struct(aa,aa);
       Struct(aa);
       Struct();
    }
}

P>Но засада класная

struct XXX {
    XXX();
    XXX( int );
};

void foo()
{
    XXX qu( 5 ); // объявили переменную qu
    XXX ququ(); // объявили функцию ququ:)
}