multiple nonlocal variables

Hello,

I am reading a book and am curious about a statement it makes:

"Anything declared in a block that contains a nested block is nonlocal to the inner block. (Global identifiers are nonlocal with respect to all blocks in the program.)"

Code:

int a = 1; //global, nonlocal

void myFunction() {
   int a = 2; //local, hides global a

   while (a < 3) {
      a++; //a is nonlocal here?
   }
}

I always thought of "a" declared within myFunction as local to the function. But the book suggests that within the while statement it is nonlocal. Is this correct? How does C++ see a difference between two nonlocal variables? What exactly does C++ do when "name hiding"?

EDIT: Just keeps reading backwards until it finds one.

Ok, so:
1. int a =1 : This is declared outside of any block and thus has file/global scope.

2. int a=2 in myFunction:

This is declared inside the block (the function), thus has block scope which terminates at the end of the block - the end of the function in this case. Since the global and local variables have the same name, the inner scope (function scope) is visible and the outer scope (global) is hidden.

3. The while loop: a is technically not local in the while block, however since the while block is inside the function block, a is visible to the while block since a has block scope for the entire function. You could technically declare another variable named a inside the while loop and it would be visible and hide the a declared by the function.

A quick example program to demonstrate the rules of scope:

Code:

#include <iostream>

int a = 2;

void foo(void);

int main(void){

	std::cout <<"a in main is: " <<a <<std::endl;
	foo();
	std::cout<<"a after function call: " <<a<<std::endl;
	std::cin.get();
	return(0);
}
void foo(void){
	int a = 3;
	int b=4;
	std::cout<<"a in foo is: " <<a <<std::endl;
	while(b<5){
		int a=5;
		std::cout<<"a in while is: "<<a <<std::endl;
		b++;
	}
	std::cout<<"a after while is: "<<a<<std::endl;
}

First of all, thank you for replying. I appreciate you taking time to answer my post and I do mean that. So I hope nobody takes this the wrong way, it is meant to be constructive criticism with a hint of humour:

What was the question as written:
"Is this right?"

What you interpreted it to be:
"How does scope work?"

What it was:
an insignificant double checking that I hadn't made some clearly obvious misreading of the text, in fact not a real question.

What you gave as answer:
160 word answer.

What would have been a fitting answer:
"Yes."

What I wanted to know:
"How does C++ see a difference between two nonlocal variables? What exactly does C++ do when "name hiding"?"

I believe I have answered the first of those two questions myself. I am still curious about how "name hiding" works. And I don't mean what it does, I understand its function, I'm wondering about the implementation.

Originally Posted by Furious5k

First of all, thank you for replying. I appreciate you taking time to answer my post and I do mean that. So I hope nobody takes this the wrong way, it is meant to be constructive criticism with a hint of humour:

What was the question as written:
"Is this right?"

What you interpreted it to be:
"How does scope work?"

What it was:
an insignificant double checking that I hadn't made some clearly obvious misreading of the text, in fact not a real question.

What it was: You expressing your non-clear understanding of scope.
What I did: Explain in clear terms what the standard says, not rewording of your book.

Originally Posted by Furious5k

What you gave as answer:
160 word answer.

What would have been a fitting answer:
"Yes."

See post above.

What I wanted to know:

Originally Posted by Furious5k

"How does C++ see a difference between two nonlocal variables?

What I did: Answered the question with my first post. Note the use of the word block scope.
What you did: Not bother to read and apply my answer to your question.

Originally Posted by Furious5k

What exactly does C++ do when "name hiding"?"

What every compiler does when resolving scope. The variable names you use are meaningless to the compiler. The process of compilation is a multiple step process. Part of that process is building a symbol tree in which the compiler will resolve block scope and thus it will know whether or not a particular variable is "in scope", e.g. can be seen.

What I did: Just answered your question, however I ensured it was short for your convenience.
What you are going to do: Not understand my answer, and then later decide that you already know the answer to this.

What it was: You expressing your non-clear understanding of scope.

Could you please explain what part of scope I am unclear on?

What you are going to do: Not understand my answer, and then later decide that you already know the answer to this.

Even assuming I was absolutely wrong about everything I said, an honest attempt to point out another person's (perceived) mistake, probably doesn't deserve an insult in return.

Originally Posted by Furious5k

Could you please explain what part of scope I am unclear on?

You asked the question

How does C++ see a difference between two nonlocal variables?

which ... well, I still don't know what you actually want to know to answer that question. (The two variables named a differ in scope and are therefore different variables, so that's how the compiler tells them apart, I guess, maybe?)

A local variable name hides any nonlocal variable of the same name. So, if you have a local variable you use it, if not you use a nonlocal one.

I interpreted this as being key to how the compiler decides on which variable to use. What confused me was that when you enter a new block, variables declared in previous block become nonlocal and this allows for the situation where you have two nonlocal declarations and no local ones. If it were a simple case of "use nonlocal" since local doesn't exist, the two declarations would be equivalent.

So the real confusion stemmed from misunderstand that a variable being local and a variable having local scope are not tied. A variable can become nonlocal and still have local scope.

I always thought of "a" declared within myFunction as local to the function.

The book simply introduced the concept of a variable being local/nonlocal, which conflicted with the meaning I previously associated with a variable being local, which was a variable having local scope.

Originally Posted by Furious5k

A local variable name hides any nonlocal variable of the same name. So, if you have a local variable you use it, if not you use a nonlocal one.

Careful with this statement....

Originally Posted by Furious5k

I interpreted this as being key to how the compiler decides on which variable to use.

Not quite.

Originally Posted by Furious5k

What confused me was that when you enter a new block, variables declared in previous block become nonlocal and this allows for the situation where you have two nonlocal declarations and no local ones. If it were a simple case of "use nonlocal" since local doesn't exist, the two declarations would be equivalent.

This is because this is not how it works.

Originally Posted by Furious5k

So the real confusion stemmed from misunderstand that a variable being local and a variable having local scope are not tied. A variable can become nonlocal and still have local scope.

Both statements are wrong. A local variable by definition has local scope, a non-local variable by definition cannot have local scope.

Originally Posted by Furious5k

The book simply introduced the concept of a variable being local/nonlocal, which conflicted with the meaning I previously associated with a variable being local, which was a variable having local scope.

Which is why I initially made my "long winded" response to you.....

There I kept my answers short this time.

Careful with this statement....

That was the best way I could type it which wasn't untrue and reflected the way I was thinking about it earlier, though I was reading too much into it at the time.

Not quite.
This is because this is not how it works.

I am aware. I was using past tense in sentences which were meant to represent what I thought before, to explain the context and hence meaning of my question.

Both statements are wrong. A local variable by definition has local scope, a non-local variable by definition cannot have local scope.

I don't understand. "a" has local scope within the whole of the function body, inclusive of the while loop, but within the loop it is "nonlocal"? So isn't there a point where it both has local scope and is "nonlocal"?

Code:

void myFunction() {
   int a = 2; //variable is local, has local scope

   while (a < 3) {
      a++; //variable is nonlocal, has local scope?
   }
}

Which is why I initially made my "long winded" response to you.....

There I kept my answers short this time.

It would be nice if you didn't belabour this.
Also, I still feel you answered the difference between global and block scope variables, not two nonlocal ones neither of which needs to be global according to the quoted text.

Code:

int a; //Let us call this a1, at file scope

int main() {
    int a; //Let us call this a2; it is local, in local scope.  a1 is not in scope
    while (rand() < 10000) {
        a = a + 2; //This uses a2; it is nonlocal to this block, but it has local scope and is currently the only a in scope
        int a; //Let us call this a3; it is local, in local scope.  a2 is now not in scope, let alone in local scope
        a = a + 2; //This must use a3, as it is the only a in scope.
    }  //a3 no longer in scope, a2 once again in scope
    a = a + 2; //This must use a2, as it is the only a in scope.
}
//Poor a1, never used

I think that covers the interesting cases for the question.

Hehe, I like the last code comment.

A local variable by definition has local scope, a non-local variable by definition cannot have local scope.

So is this untrue then? Or is it a true definition and this is simply a pedantical technicality under which it doesn't hold up? or what?

Originally Posted by Furious5k

So is this untrue then? Or is it a true definition and this is simply a pedantical technicality under which it doesn't hold up? or what?

It is very true. a2 in tabstop's example has function scope, which means it remains visible for the entire function, it is not local to the inner block (the while loop) although it is visible until the declaration of a3.

Tabstop's example is illustrative, though I wish he would do it with well defined code. The result of a+2 when a is uninitialized is undefined. Furthermore, on some, admittedly rare systems a+2 could result in an integer overflow causing an inturupt or structural exception to effectively crash the program.

It would probably be simpler to illustrate the point by modifying tabstop's sample so it prints out the address of a in various places, rather than trying to access or modify its value.

From a couple of sources:
"The only type of identifier with function scope is a label name." I'm not sure why this term is now being used.

This is how I'm understanding what you mean by function scope:
Function scope has the same meaning as local/block scope, except you have to be talking about a function block when using the term, no? So not all local scope is function scope, but all function scope is also local scope.

a non-local variable by definition cannot have local scope

a = a + 2; //This uses a2; it is nonlocal to this block, but it has local scope

The way I read it, you seem to be saying a2 doesn't have local scope, rather having function scope. But searching for function scope gave me no definition to support this.