Thread: How is this possible?

Is this not true?

Yes:

Code:

int x = 0;

if (true) {
  x = 12345;
}

cout << x; // Prints 12345

But that's not true if you define a new variable in the nested scope with the same name. Then the inner name hides the outer name and changes made are to the inner name, which is destroyed when execution leaves the nested scope:

Code:

int x = 0;

if (true) {
  int x = 12345;
}

cout << x; // Prints 0

Is it true in all cases that a variable defined outside a scope and used inside a sub-scope will, after the sub-scope has been executed, retain the value that was set to it, if there was any change, in the sub-scope?

Well, technically, yes. But apparently no, because the sub scope could be function, and by default, C++ is pass-by-value. In that case you'd make a copy, edit the copy, and the changes made won't be mirrored in the calling function. You can use a reference or a pointer to fix that though.

Code:

int factorial = 1;
for (int x = 2; x <= num; ++x)
   factorial = factorial * x;
cout << num <<"! = " << factorial <<"\n";

//Since factorial is declared outside of the for-loop, it is still in scope here.
//It will stay in scope until a } is encountered, probably at the end of this function.
//This means any changes made to it are kept until a } is encountered.

it wouldn't have any different effect as an end result, if there is any difference it would be that ++ and -- compile to increment and decrement instructions. (most archs support them) instead of addition and subtraction instructions.

++ and -- only for convenience.

edit: There is one thing you can do, the pre-increment and decrement operators for example:
boat = theJunk[++i] ;
would first increment i then set boat to theJunk[i] while:
boat = theJunk[i++]
would set boat to theJunk[i] and then increment i

A decent optimising compiler would convert i = 1 + 1; to an INC op anyway.