Out of Scope?

[XianForce]

Well I was reading my Sams Teach Yourself C++ in an Hour a Day and one of the questions asked this:

What is the value of x when the for loop completes?

Code: Select all

for (int x = 0; x < 100; x++)

I thought it would be 100, because the first statement it defines and initializes the variable, so I would think it should be 100.

The answer in the book says: "The variable is out of scope and thus has no valid value"

I'm a little confused by this. Is this just meaning after the for loop completely ends? or the final value of x?

My best guess is the variable was created within the loop so once the loop ends, its out of scope, and the question is referring to after the loop ends.

[avansc]

yes, any variables inside loops are just local to that loop

so if you were to do

for(int a = 0;a< 100;a++)
{
}

int x = a;

you would get an error.

however

int a = 0;
for(a;a<100;a++)
{
}
int x = a;

that will work.

[XianForce]

yes, any variables inside loops are just local to that loop

so if you were to do

for(int a = 0;a< 100;a++)
{
}

int x = a;

you would get an error.

however

int a = 0;
for(a;a<100;a++)
{
}
int x = a;

that will work.

Ok, so my last statement was right?

[avansc]

yup

[XianForce]

yup

Alright thank you

[MarauderIIC]

Just for fun...

Code: Select all

int a = 100;
{
    int a = 50;
    cout << a << endl;
}
cout << a << endl; //this a is not that a :)

[XianForce]

Just for fun...

Code: Select all

int a = 100;
{
    int a = 50;
    cout << a << endl;
}
cout << a << endl; //this a is not that a :)

Easy, there are 2 variables named a with this one. You declared a new a within the brackets so it would display 50, but after the brackets end, that a would go out of scope. Thus, a would still equal 100.