6.9. Local variables¶

About this time, you might be wondering how we can use the same variable i in both printMultiples and printMultTable. Didn’t I say that you can only declare a variable once? And doesn’t it cause problems when one of the functions changes the value of the variable?

The answer to both questions is “no,” because the i in printMultiples and the i in printMultTable are not the same variable. They have the same name, but they do not refer to the same storage location, and changing the value of one of them has no effect on the other.

Note

Remember that variables that are declared inside a function definition are local. You cannot access a local variable from outside its “home” function, and you are free to have multiple variables with the same name, as long as they are not in the same function scope.

The stack diagram for this program shows clearly that the two variables named i are not in the same storage location. They can have different values, and changing one does not affect the other.

Notice that the value of the parameter n in printMultiples has to be the same as the value of i in printMultTable. On the other hand, the value of i in printMultiples goes from 1 up to n. In the diagram, it happens to be 3. The next time through the loop it will be 4.

It is often a good idea to use different variable names in different functions, to avoid confusion, but there are good reasons to reuse names. For example, it is common to use the names i, j and k as loop variables. If you avoid using them in one function just because you used them somewhere else, you will probably make the program harder to read.

Q-1: Are there any issues with the code below?

#include <iostream>
using namespace std;

int main() {
  cout << "Let's print the multiples of 2." << endl;
  int i = 1;
  while (i < 10) {
    int j = i * 2;
    cout << i << ": " << j << endl;
    i++;
  }
  i = 10;
  j = 20;
  cout << i << ": " << j << "!";
}

Yes, we cannot output the value of j outside of the loop.
The scope of i is restricted to the loop, so we cannot change the value of i outside of the loop.
Yes, we cannot output anything before the loop.
This is allowed.
Yes, we cannot reassign j to 20 outside of the loop.
The scope of i is restricted to the loop, so we cannot output the value of i outside of the loop.
Yes, we cannot let i start at 1 in the loop.
We are allowed to initialize i to any value.
No, there are no issues with the code below.
There are issues with the code. Can you find them?

Q-2: Take a look at the code below. Is the i in printMultiples the same variable as the i in printMultTable?

#include <iostream>
using namespace std;

void printMultiples (int n) {
  int i = 1;
  while (i <= 6) {
    cout << n * i << '\t';
    i = i + 1;
  }
  cout << endl;
}

void printMultTable() {
  int i = 1;
  while (i <= 6) {
    printMultiples (i);
    i = i + 1;
  }
}

int main() {
  printMultTable();
}

Yes
They are two different variables in two different scopes, but they do have the same name.
No
Correct! They are not the same variable.

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Q-3: Take a look at the code below. Is the variable j accessable in the function printMultiples?

#include <iostream>
using namespace std;

void printMultiples (int n) {
  int i = 1;
  while (i <= 6) {
    cout << n * i << '\t';
    i = i + 1;
  }
  cout << endl;
}

void printMultTable() {
  int j = 1;
  while (j <= 6) {
    printMultiples (j);
    j = j + 1;
  }
}

int main() {
  printMultTable();
}

Yes
The scope of j does not include printMultiples function.
No
Correct! j is not accessable as the value is merely passes from one function to another. We cannot have a statement such as j++; in printMultiples as it is out of the scope of printMultTable

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