// Sample (tested!) code to illustrate the ideas.#include <iostream> // for input and output#include <string> // for the std::string typeusing namespace std; // tells the compiler to look in the std namespace// for any identifiers (global variables, type names, etc.) that it can't// find definitions for in our own code.int main() {  // In C++, main() should return int, never void.  string message = "hi "; // we can initialize a string using a char*  message += "there!"; // adding char* to a string works too  3 + 4; // a computation on its own is valid, but won't "do anything" unless  // you store the result somewhere or otherwise cause a side effect (like  // outputting the result). I'm just doing this to show:  // "hi " + "there!"; <-- not valid! Both pointers; can't add them.  // Anyway, a good compiler will give you a warning for doing something like  // this (or can at least be configured to do so), but will still compile your  // code.  cout << message << endl; // output the message.  char x = 8;  int y = 3;  // Notice I declared y as a char type; this is to show that the individual  // "char"acters of a char* are actually numeric values - the console maps  // them into display characters via ASCII.  // Output the division as a fraction.  // We'll do this by showing the whole number part, then a space, then  // the remainder, then a slash, then the divisor.  cout << "8/3 = " << (x/y) << ' ' << (x%y) << '/' << y << endl;  // Note that *single* quotes produce a character literal, not a string  // literal. Thus it can only hold one byte, but it's not a pointer, it's  // an integral type - so you can do arithmetic with it. Now, to do  // *meaningful* arithmetic is the hard part ;) There's really only one  // trick you are likely to rely on there:  cout << static_cast<char>('0' + x) << endl;  // Here we sum '0' - which is 48, because of that encoding - and 8, to get 56  // - which is '8' as a printable character. The trick is that digits '0'   // through '9' are sequential in ASCII. However, to print the character '8'  // rather than the numeric value 56, we need to "cast" the result back to  // a char type - because summing the two chars actually gave us an int.  // To finish off, let's show the division as a floating-point value. We'll  // cast x to a floating-point type, in this case "double". Division between  // "mixed" types (one floating-point and one integral) gives a floating-point  // result, according to the rules of the language.  cout << "8/3 = " << (static_cast<double>(x) / y) << endl;}