Scope of variables in if statements - c++

Question

I have a class that has no default constructor or assignment operator so it is declared and initialized within an if/else statement depending on the result of another function. But then it says that it is out of scope later even though both routes of the conditional will create an instance.
Consider the following example (done with int just to illustrate the point):
#include <iostream>
int main()
{
if(1) {
int i = 5;
} else {
int i = 0;
}
std::cout << i << std::endl;
return 0;
}
Do variables declared in a conditional go out of scope at the end of the conditional? What is the correct way to handle the situation where there is no default constructor but the arguments for the constructor depend on certain conditionals?
Edit
In light of the answers given, the situation is more complex so maybe the approach would have to change. There is an abstract base class A and two classes B and C that derive from A. How would something like this:
if(condition) {
B obj(args);
} else {
C obj(args);
}
change the approach? Since A is abstract, I couldn't just declare A* obj and create the appropriate type with new.

Answer 1

"Do variables declared in a conditional go out of scope at the end of the conditional?"
Yes - the scope of a local variable only falls within enclosing brackets:
{
int x; //scope begins
//...
}//scope ends
//x is not available here
In your case, say you have class A.
If you're not dealing with pointers:
A a( condition ? 1 : 2 );
or if you're using a different constructor prototype:
A a = condition ? A(1) : A(2,3);
If you're creating the instance on the heap:
A* instance = NULL;
if ( condition )
{
instance = new A(1);
}
else
{
instance = new A(2);
}
or you could use the ternary operator:
//if condition is true, call A(1), otherwise A(2)
A* instance = new A( condition ? 1 : 2 );
EDIT:
Yes you could:
A* x = NULL; //pointer to abstract class - it works
if ( condition )
x = new B();
else
x = new C();
EDIT:
It seems what you're looking for is the factory pattern (look it up):
class A; //abstract
class B : public A;
class C : public A;
class AFactory
{
public:
A* create(int x)
{
if ( x == 0 )
return new B;
if ( x == 1 )
return new C;
return NULL;
}
};

Answer 2

Do variables declared in a conditional go out of scope at the end of
the conditional?
Yes.
What is the correct way to handle the situation where there is no
default constructor but the arguments for the constructor depend on
certain conditionals?
Write a function that returns a value, from which you copy.
T foo()
{
if(condition)
return T(x);
return T(y);
}
void bar()
{
T i(foo());
}
Edit:
Since A is abstract, I couldn't just declare A* obj and create the
appropriate type with new.
What do you mean? That's exactly how dynamic typing works. Except I wouldn't use a raw pointer, I would use a unique_ptr.
std::unique_ptr<A> obj;
if(condition) {
obj = std::unique_ptr<A>(new B(args));
} else {
obj = std::unique_ptr<A>(new C(args));
}

Answer 3

Your alternative will be pointers:
MyObject *obj;
if(cond1)
{
obj = new MyObject(1, 2, 3);
}
else
{
obj = new MyObject(4, 5);
}
Remember to delete it when you are done with it, or use a smart pointer.

Answer 4

Yes it will be out of scope if declared in a conditional, loop etc. Will the type of the variable change depending on the conditional?