lect09, Thu 05/02

Passing Parameters to functions; Numerical base conversions

From now on, include pre- and post- conditions for the functions that you write: * Precondition: What you expect the function to receive * Poscondition: What you expect will be true once the function has completed * These should be added as a pair of comments right before each function * If you update your function, make sure to read over your pre- and post- conditions to make sure they still make sense and match the new code. Update them if necessary as you edit your code.

Code from Lecture

First, we wrote a simple function to swap the values of two variables. This example motivates the need for the pass-by-reference mechanism, since we cannot return more than one value from a function.

// Lecture 9 functions
// Pass-by-value vs. pass-by-reference

#include <iostream>
using namespace std;

void swap(int& val1, int& val2);
// A function to swap two integers
// Precondition: val1 and val2 are integers
// Postcondition: val1 has the value of val2
// and vice versa

int main()
{
    //int size = 3;
    //int arr[size];
    /* Declaring a function inside of another function leads to an error
void swap1(int& val1, int& val2)
{
    cout << "Hi from swap1";
}
*/

    int num1, num2;
    cout << "Enter 2 numbers: ";
    cin >> num1 >> num2;
    cout << "Num1 = " << num1 << endl;
    cout << "Num2 = " << num2 << endl;

    swap(num1, num2);
    //swap1(num1, num2);

    cout << "Num1 = " << num1 << endl;
    cout << "Num2 = " << num2 << endl;

    return 0;
}
void swap(int& val1, int& val2)
// A function to swap two integers
// Precondition: val1 and val2 are integers
// Postcondition: val1 has the value of val2
// and vice versa
{
    int temp = val1;
    cout << "---" << endl;
    cout << "Val1 = " << val1 << endl;
    cout << "Val2 = " << val2 << endl;
    val1 = val2;
    val2 = temp;
    cout << "Val1 = " << val1 << endl;
    cout << "Val2 = " << val2 << endl;
    cout << "---" << endl;
    //main();
}

We then use the & symbol that allows us to print the addresses of the variables to see the effect of pass-by-value (the values are copied into a different address) vs. pass-by-reference (the function uses the values stored at the same address that they were in the main function).

// Lecture 9 functions

#include <iostream>
using namespace std;

//void swap(int &val1, int &val2);
void swap(int val1, int val2);
// A function to swap two integers
// Precondition: val1 and val2 are integers
// Postcondition: val1 has the value of val2
// and vice versa

int main()
{
    int num1, num2;
    cout << "Enter 2 numbers: ";
    cin >> num1 >> num2;
    cout << "Num1 = " << num1 << endl;
    cout << "Num2 = " << num2 << endl;
    cout << "num2& = " << &num1 << endl;
    cout << "Num2& = " << &num2 << endl;

    swap(num1, num2);  // The address locations change, depending on the formal parameters

    cout << "Num1 = " << num1 << endl;
    cout << "Num2 = " << num2 << endl;
    cout << "num2& = " << &num1 << endl;
    cout << "Num2& = " << &num2 << endl;

    return 0;
}
//void swap(int &val1, int &val2)
void swap(int val1, int val2)
// A function to swap two integers
// Precondition: val1 and val2 are integers
// Postcondition: val1 has the value of val2
// and vice versa
{
    int temp = val1;
    cout << "---" << endl;
    cout << "Val1 = " << val1 << endl;
    cout << "Val2 = " << val2 << endl;
    val1 = val2;
    val2 = temp;
    cout << "Val1 = " << val1 << endl;
    cout << "Val2 = " << val2 << endl;
    cout << "Val1& = " << &val1 << endl;
    cout << "Val2& = " << &val2 << endl;
    cout << "---" << endl;
    //main(); // fun test to see the infinite recursion
}

## Counting Numbers in Different Numerical Bases:

• Computers count in base-2 (binary→ 0s and 1s), but we count in base-10 (decimal→ 0-9). This is because bits on hardware are easily represented in binary- they are tiny switches that are either on or off.
• Other 2-related bases for counting in are:
  • Octal- base-8→ 3 bits (2³ = 8) (0-7)
  • Hexadecimal- base-16 → 4 bits (2⁴ = 16) (0-9, A-F)
    • A=10, B=11, C=12, D=13, E=14, F=15

Positional Notation

• In Decimal:
  • Each position corresponds to a different power of 10
    • 642 = (6*10²) + (4*10¹) + (2*10⁰)
• Using the same principle, we can convert from any base into decimal
  • What is 642 in base-13??
  • Each position is a power of 13
  • (6*13²)+(4*13¹)+(2*10⁰) = 1063

Converting Binary to Decimal

• What is 1011 (base-2) in decimal?
  • Start with the rightmost digit, 2⁽⁰⁾, and move up, multiplying by a larger power of 2 every time.
    • 1*(2³) + 0*(2²) + 1*(2^¹) + 1*(2⁰) = 8 + 0 + 2 + 1 = 11
• What if you add a bit to most significant end?
  • Now the binary number is: 11011 (which is 10000 + 1011 in binary)
  • Perform the same operations, but add 1*(2⁴) = 16
    • 16+11 = 27

Converting Binary to Octal or Hexadecimal

• Group the bits!
  • For octal, group the bits in the binary number by 3, starting from least significant (far right)
    • Because binary is base-2, and octal is base-8, which is 2³
  • For hex, group by 4s from far right
    • Because binary is base-2, and hex is base-16, which is 2⁴
  • Examples on slides!

Algorithm for converting Base-10 to other bases

• Available on slides in much greater detail (with examples)
  • Overview for binary example:
    • Divide decimal number by 2- the remainder will be 0 or 1
    • Add this to the left as the most significant bit of the the binary number you are producing
    • Divide the remaining “half” by 2, repeat w remainder, etc until remainder is 0
    • The string of 1s and 0s you have produced from the remainders is the binary conversion
  • This process can be done with any number base- just divide by that base number repeatedly, rather than by 2

Why is this useful?

• Basics of memory manipulation and understanding
• This is how everything is represented in memory
• “Reverse engineering” of binary representations of objects
• Pointers access, use, trade memory addresses
• Memory addresses are hex values
• You will be using hexadecimals when you assign pointers.

Practice Questions

1. What does the following code print out?

   void foo(int& x){
   x++;
   }
   int main(){
   int a = 5;
   foo(a);
   cout << a << endl;
   return 0;
   }
   

2. What does the following code print out? Assume it’s part of a working C++ program

   int x = 0;
   int& y = x;
   y = y + 20;
   x = x - 15;
   cout << x << endl;
   cout << y << endl;
   

3. Convert the decimal number 166 into hexadecimal and binary