
△ABC
= ADEC - △ADB - △BEC
= 1/2(x1y2 + x2y3 + x3y1 - x2y1 - x3y2 - x1y3)
=

2. Heron's Formula
△= sqrt(s(s-a)(s-b)(s-c))
x = 1/2 (a+b+c)
Ref.:
http://en.wikipedia.org/wiki/Triangle
http://euler.tn.edu.tw/area.pdf



myvar = 25; foo = &myvar; bar = myvar;2. Dereference Operator (*)
baz = *foo;
#include <iostream>
using namespace std;
int main ()
{
int numbers[5];
int * p;
p = numbers;
*p = 10;
p++;
*p = 20;
p = &numbers[2];
*p = 30;
p = numbers + 3;
*p = 40;
p = numbers;
*(p+4) = 50;
for (int n=0; n<5; n++)
cout << numbers[n] << ", ";
return 0;
Output:10, 20, 30, 40, 50,
int x;
int * p1 = &x; // non-const pointer to non-const int
const int * p2 = &x; // non-const pointer to const int
int const * p3 = &x; // also non-const pointer to const int
int * const p4 = &x; // const pointer to non-const int
const int * const p5 = &x; // const pointer to const int
char a; char * b; char ** c; a = 'z'; b = &a; c = &b;
#include <iostream>
using namespace std;
int addition (int a, int b)
{ return (a+b); }
int subtraction (int a, int b)
{ return (a-b); }
int operation (int x, int y, int (*functocall)(int,int))
{
int g;
g = (*functocall)(x,y);
return (g);
}
int main ()
{
int m,n;
int (*minus)(int,int) = subtraction;
m = operation (7, 5, addition);
n = operation (20, m, minus);
cout << n;
return 0;
}
Output:
8Mostly, we use function point in sqort().
(*p).foo; p->foo;