What are the implications of misidentifying a local and global variable?

Answer 1

2 answers:

1. You made a variable global; it should have been local.

No biggie. Although deemed 'unprofessional' by most programmers, nothing can really go wrong if you make global variables.

2. You made a local variable instead of a global one.

Big issue. If you define a variable IN a function, then the variable is deleted when the function ends. This means that when you call the function again, that variable will act exactly the same as it did last time. A global variable will keep its value stored until the end of the program, which means it can change during the course of the program.

Suppose you do this:

void printCounter()

{

int i = 1;

cout << i; //this is C++ but this basically prints the value of i onscreen.

i++;

cout << i;

}

This is a local variable.

If you run this twice, you will get the output

1 2 1 2

Suppose you do this:

int i =1;

void printCounter()

{

cout << i; //this is C++ but this basically prints the value of i onscreen.

i++;

cout << i;

i++;

}

Now the value of i is saved, even when the first printCounter function ends.

Your output will be:

1 2 3 4

Answer 2

something bad happens... like getting it wrong on your ITT capstone assessment questionnaire...

this answer is awesome! and correct.

Answer 3

Either compilation or run-time error. The results are unpredictable.