I want to declare multiple objects of the same type and initialize them with the same rvalue by just one expression; without the need to declare and initialize them by separate statements.
What I want is something like:
int a = b = 10; // dummy-statement. This does not work.
or
int a = int b = 10; // dummy-statement. This does not work either.
instead of
int b = 10;
int a = b;
Is there a way to do that?
Technically, yes: int a = value, b = a;
, or you might consider int a, b = a = value;
. Without repeating identifiers, no, at least not in C; the grammar simply does not provide for it. Each ”declarator =
initializer” in the grammar can declare only one object, per grammar production in C 2018 6.7.6 1 and explicit statement in 6.7.6 2: “Each declarator declares one identifier…”
As mentioned in a comment, here is a horrible way to do it in C++. I make no representations regarding C++’s rules about order of initialization in a single declaration, issues about threading, and so on. This is presented as an educational exercise only. Never use this in production code.
template<class T> class Sticky
{
private:
static T LastInitializer; // To remember last explicit initializer.
T Value; // Actual value of this object.
public:
// Construct from explicit initializer.
Sticky<T>(T InitialValue) : Value(InitialValue)
{ LastInitializer = InitialValue; }
// Construct without initializer.
Sticky<T>() : Value(LastInitializer) {}
// Act as a T by returning const and non-const references to the value.
operator const T &() const { return this->Value; }
operator T &() { return this->Value; }
};
template<class T> T Sticky<T>::LastInitializer;
#include <iostream>
int main(void)
{
Sticky<int> a = 3, b, c = 15, d;
std::cout << "a = " << a << ".\n";
std::cout << "b = " << b << ".\n";
std::cout << "c = " << c << ".\n";
std::cout << "d = " << d << ".\n";
b = 4;
std::cout << "b = " << b << ".\n";
std::cout << "a = " << a << ".\n";
}
Output:
a = 3. b = 3. c = 15. d = 15. b = 4. a = 3.