Scope resolution operator

[Shan]

In the following code:

func()
{
int SmeVar;
{
int SmeVar;
//Here I want to use SmeVar defined in the outer block how
can I do that ?
}
}

 

[Jerry Coffin]

(e-mail address removed)>, (e-mail address removed)
says...

In the following code:

func()
{
int SmeVar;
{
int SmeVar;
//Here I want to use SmeVar defined in the outer block how
can I do that ?
}
}

It's not possible unless the namespace in which SmeVar is located has a
name, which it doesn't in this case.

 

[Stefan Naewe]

In the following code:

func()
{
int SmeVar;
{

int& outer_SmeVar = SmeVar;

int SmeVar;
//Here I want to use SmeVar defined in the outer block how
can I do that ?
}
}

S.

 

[Shan]

int& outer_SmeVar = SmeVar;


S.

Hi,
Thanks for the reply, can you suggest any way of doing this using
scope resolution operator?

 

[Stefan Naewe]

Hi,
Thanks for the reply, can you suggest any way of doing this using
scope resolution operator?

No.
Read what Jerry said. You need a named scope.

S.

 

[Jim Langston]

Hi,
Thanks for the reply, can you suggest any way of doing this using
scope resolution operator?

Sure. Give one of the variables a different name.