// solution to lab8 problem 2

#include<iostream.h>

// write a function f that takes an integer x and return an integer
// representing 3*x + 1.

int f(int x)
{ return 3*x + 1; }

// write a function s that takes an integer x and returns no value.
// the function's code should first change the value of x to x+1, then
// print out the new value of x. e.g., s(3) should print 4.

void s(int x)
{ x = x+1;
  cout << "local x is now " << x << "\n";
}

// write a function g that takes an integer n and a character.  The
// character is expected to be 'd' or 's'.  If it's 'd', return 2*n,
// otherwise, if it's 's' return n*n.  If it's a letter that's not 'd' or
// 's', return 0.

int g(int n, char c)
{ if ( c == 'd') return 2*n;

  if (c == 's') return n*n;
      else return 0;
  // end outer else 
}
// write a function allthree that takes in three 'bool' parameters and returns
// true if all three boolean values are true, and returns false otherwise.
// For example, allthree(3<4, 5==2+3, 6>1) should return true.

bool allthree(bool a, bool b, bool c)
{ return a && b && c; }

// complete the following main function:
int main()
{
  // declare variables x, and y, and assign them 3 and 4 respectively
  int x;
  int y;
  x = 3;
  y = 4;
  // declare a bool variable b;
  bool b;
  // assign y to the result of f applied to x;
  y = f(x);
  // call function s on x;  -- note this is a statement, since s returns void
  s(x);
  // print out the value of x.  Why did it print 3?  Wasn't it changed to
  // 4 inside the function s?!  You need to understand what's happening 
  // here (local variables).
  cout << "x in main is still " << x << "\n";  

  // assign to x the result of calling g on y and 's'.
  x = g(y,'s');

  // call the allthree function on the following boolean test values:
  //  x is greater than 2
  //  y is greater x
  //  y is less than 15
  // assign the value returned by the function call to b.
  b = allthree( (x>2), (y>x), (y<15) );

  // Print out the final values of x, y, and b, indicating which is which.
  cout << "x==" << x << " y==" << y << " b==" << b << "\n";
 return 0;
}