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

Question

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What is the scope of a 'while' and 'for' loop? (10 answers)

Closed 8 years ago.

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

Answer 1

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.

Answer 2

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

Answer 3

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