Added materail for Section08 thus far: Assignment, Arithmetic, and Euros.

Author: anthonyclyan

Section08-Statements_and_Operators/README.md

@@ -79,3 +79,11 @@
 - assignment expressions is evaluated to what was just assigned
 - more than one variable can be assigned in a single statement
 
+## Arithmetic Operators
+
+- Addition (`+`)
+- Subtraction (`-`)
+- Multiplication (`*`)
+- Division (`/`)
+- Modulo or remainder (`%`)
+    - only works with integers

Section08-Statements_and_Operators/section8_source_code/ArithmeticOperators/main.cpp

@@ -19,7 +19,7 @@ int main() {
     int num1 {200};
     int num2 {100};
 
-  //  cout << num1 << " + " <<  num2 << " = "<< num1+ num2 << endl;
+    cout << num1 << " + " <<  num2 << " = "<< num1+ num2 << endl;
 
     int result {0};
 
@@ -37,21 +37,32 @@ int main() {
 
     result = num2 / num1;
     cout << num2 << " / " <<  num1 << " = "<< result << endl;
-    
+    // since we are working with integers the 5 in 0.5 will be chopped and only 0 is outputted
+    // use double instead of integer type
+
     result = num1 % num2;
-    cout << num1 << " % " <<  num2 << " = "<< result << endl;    
+    cout << num1 << " % " <<  num2 << " = "<< result << endl;  
+    // 200 divided by 100 = 2 with no reamainder, remainder = 0, result = 0
 
     num1 = 10;
     num2 = 3;
 
     result = num1 % num2;
-    cout << num1 << " % " <<  num2 << " = "<< result << endl;    
+    cout << num1 << " % " <<  num2 << " = "<< result << endl;  
+    // 10 divided by 3 = 9 with reaminader = 1, result = 1
 
-    result = num1 * 10 + num2;
+    result = num1 * 10 + num2;  
+    // compiler might read right to left but arithmetic operations still holds true
+    // num1 will be multiplied by 10 first, then add num2
+
+    // if one desire to multiply num1 by 10 + num2, then we need perentesis ()
+    result = num1 * (10 + num2);
 
     cout << 5/10 << endl;
+    // output 0 since 0.5 is not an integer, the 5 will be chopped
 
     cout << 5.0/10.0 << endl;
+    // output 0.5 since we are using double instead of integer with the .0
 
     cout << endl;
     return 0;

Section08-Statements_and_Operators/section8_source_code/AssignmentOperator/main.cpp

@@ -7,15 +7,37 @@ using namespace std;
 
 int main() {
 
+    // variables initialization
     int num1 {10};
     int num2 {20};
 
-    num = 100;
-    
+    // assignment statement: 
+    num1 = 100;          // assign 100 into num1
+    // num1 = num2;         // assign num2 value into num1
+    // num1 = num2 = 1000;  // works since compiler reads from right to left
+    // num1 = "Frank";      // compiler will try to put string type into int type but will fail and output "conversion error"
+
     cout << "num1 is " << num1 << endl;
     cout << "num2 is " << num2 << endl;
 
     cout << endl;
     return 0;
 }
 
+/*
+int main() {
+    
+    // variables initialization
+    const int num1 {10};
+    int num2 {20};
+    
+    // num1 = 100;  // this doesn't work because num1 is a constant, compiler will throw "read-only variable error"
+    // 100 = num1;  // compiler doesn't understand and throw error "lvalue required as left operand of assignment"
+    
+    cout << "num1 is " << num1 << endl;
+    cout << "num2 is " << num2 << endl;
+
+    cout << endl;
+    return 0;
+}
+*/

Section08-Statements_and_Operators/section8_source_code/Euros/main.cpp

@@ -10,7 +10,7 @@ int main() {
     const double usd_per_eur {1.19};
 
     cout << "Welcome to the EUR to USD converter" << endl;
-    cout << "Enter the value in EUR : ";
+    cout << "Enter the value in EUR: ";
 
     double euros {0.0};
     double dollars {0.0};