C / C++ / MFC

#include <iostream>
#include <array>
using namespace std;

constexpr size_t SIZE = 3;

class Vector
{
  array<int,SIZE> x{};

public:
  Vector (const array<int,SIZE> & a)
  {
    for (size_t n=0; n<SIZE; ++n)
      x[n] = a[n];
  }

  int operator * (const Vector & v) const
  {
    int result = 0;
    for (size_t n=0; n<SIZE; ++n)
      result += x[n] * v.x[n];
    return result;
  }
  friend ostream & operator << ( ostream & os, const Vector & v);
};

ostream & operator << ( ostream & os, const Vector & v)
{
  for (const auto & i : v.x)
    os << i << " ";
  return os;
}

int main()
{
  array<int, SIZE> x{1,2,3};
  array<int, SIZE> y{6,3,9};

  Vector v1{x};
  Vector v2{y};

  cout << "v1 " << v1 << endl;
  cout << "v2 " << v2 << endl;
  cout << "v1*v2 = " << (v1*v2) << endl;
}

Vaclav_ wrote:

char *TXBuffer = (char*) 0x04;

unsigned char TXBuffer = 0x04; // a single character
function(&TXBuffer); // use the addressOf operator to pass the address of the character to the function 

// or

unsigned char TXBuffer[] = { 0x04, 0 }; // multiple characters
function(TXBuffer); // TXBuffer is an array, so its name will be translated to its address by the compiler.


// the receiving function should be coded as
xxx function(unsigned char* buffer)
{
    // xxx is the return type
    // the function must know how many characters will be sent in the buffer
    // either a number that is always the same, or the length passed as a second parameter.
    // to access items from the buffer you use:
    unsigned char value = *buffer; // get the first value
    buffer++;                      // increment the pointer to the next element (if more than 1)
    value = *buffer;               // get the next value 
}

focusdoit wrote:

char str123[] = "Test123\r\n"; //Global
UART_print(strtest); //in a loop

focusdoit wrote:

But when I try code:

char str123[] = "Test123\r\n";  //Global
UART_print(strtest); //in a loop

Faith Burnett wrote:

To be honest I’m new to this and thought it would simply print out two right aligned pyramids with a gap in between.

for (int i = 0; i < h; i++)
{
    ...
    // Print out this many spaces
    for (int j = 0; j < h - i - 1; j++)
    ...
    // Print left hashes
    for (int j = 0; j < i + 1; j++)
    ...
    // Print right hashes
    for (int j = 0; j < i + 1; j++)
    ...
}

cout << hex << (int)buf[0] << endl;