Are variables declared in loops visible to rest of the code?

Question

I was reading this book and it said : Variables declared in a scope aren't visible outside it.

It also said : Scopes are declared by 2 curly braces - like a block of code.

So, if I had a situation like this :

for(_statement1_)
{
  int var;
  /*code*/
}
cout << var << " number of rockets left.\n";

Would the value printed be the same as the value of the var declared in the loop?

Thank you

It would be a compiler error. `var` is not declared in the outer scope. — juanchopanza, Feb 24 '14 at 06:49
Alright. So if you could write it as an answer, I'll declare the question as answered. — ABanerjee, Feb 24 '14 at 06:51
I'll pass. This is too trivial to be answered, and I am sure there are duplicates. — juanchopanza, Feb 24 '14 at 06:53
@AnitejBanerjee: But clearly you are somehow connected to the internet. Try this: https://ideone.com/ See: https://ideone.com/HSwjwU — Csq, Feb 24 '14 at 07:14

score 3 · Accepted Answer · answered Feb 24 '14 at 06:51

As your book says, the variable is scoped inside the loop's block, and isn't visible outside it.

Your code won't compile unless there is a different variable var outside the loop's scope. If there is, then the final statement would use that, not the one in the loop which is now out of scope.

score 1 · Answer 2 · answered Feb 24 '14 at 06:51

1

No, you can not use that variable outside the loop. var is visible only inside the loop.

answered Feb 24 '14 at 06:51

Chethan N

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You meant to write `can't`, right? – ABanerjee Feb 24 '14 at 06:51
yes, typing mistake initially!! – Chethan N Feb 24 '14 at 06:52

score 0 · Answer 3 · answered Feb 24 '14 at 06:56

0

It will be a compiler error since the scope of var limited inside for loop only.

answered Feb 24 '14 at 06:56

Subhajit

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Are variables declared in loops visible to rest of the code?

3 Answers3