Why does this work?

[Zyborg]

I am currently working through the book "Accelerated C++" and I've come across an example that I don't understand. The example given is supposed to make the reader have a better understanding of local variables and their lifespan within curly braces. However this one example goes against my understanding of how this works.

The example is:

int main()
{
	{ const string s = "a string";
	cout << s << endl;
	{ const string s = "another string";
	cout << s << endl; }}
	return 0;
}

In the book it says that a local variable only exists within a set of curly braces, and is destroyed once it reaches the }. In this example, const string s is initialized again within another set of curly braces before the end of the first set of curly braces. Why is it that when I compile this, it prints:

a string
another string

should this happen? Or should there be an error? If a local variable is destroyed once it reaches the last curly brace, then wouldn't the initialization of const string s in the second set of curly braces attempt to overwrite the original const string s? I am using microsoft visual studio 2010.

[filipe]

There are two different variables there. The outer s still exists when the inner s is declared. Inside the inner scope, you can't talk about the outer s, though.

[Zyborg]

Thanks filipe, that makes sense. I just tried out what you said by writing this:

int main()
{
	{string s = "a string";
	cout << s << endl;
	{string s = "another string";
	cout << s << endl;}
	cout << s << endl;}
	return 0;
}

and the result shows that the first s is left as it was.

a string
another string
a string

Can you please be more specific about what you said about how "Inside the inner scope, you can't talk about the outer s"?

this:

int main()
{
	{string s = "a string";
	{cout << s << endl;}}
	return 0;
}

still prints

a string

[filipe]

I meant in that case, where you had a local s clouding the outer one.

[Intrexa]

It keeps stepping up scopes until it finds a valid variable. Note that functions won't do that tho. It's bad practice tho to have multiple variables named the same thing and to rely on that.

[Zyborg]

Interesting. Thanks for the info Intrexa and filipe.