C / SFML:使用两个递归调用在屏幕上打印凸形形状仅显示第一个递归调用中的形状,而不是第二个

C++ / SFML: Printing convex shapes to the screen using two recursive calls only displays the shapes from the first recursive call and not the second

本文关键字:递归 调用 显示 第一个 第二个 打印 屏幕 SFML 两个      更新时间:2023-10-16

我正在使用sfml并在C 中进行编码。我正在编写的程序必须是递归实施。

我的目标是创建一个函数,该函数递归地将正方形绘制在屏幕上,并取决于先前绘制的正方形。

每个随后的正方形应小于上一个函数调用,并旋转45度向左(从上一个正方形的左角)或前一个正方形的右侧45度。

每个新广场都会产生两个正方形等。

我的想法是将左上点和正方形的上点传递到两个不同的递归功能调用,并将这些点用作后续正方形的起点。

虽然生成的正方形也将通过左上和右上角以递归功能呼叫等。

我开发的代码没有显示两个正方形,这些正方形应该是从递归函数调用中生成的。仅显示一侧。

我已经开发了以下代码(请原谅我的代码..我没有在C 中编码太久了。)

程序的驱动程序(main.cpp)

#include <SFML/System.hpp>
#include <SFML/Graphics.hpp>
#include <SFML/Window.hpp>
#include "PTree.hpp"
using namespace std;
using namespace sf;
int main( int argc, char* argv[ ] )
{
  double L = 0.0; // Length of square sides
  int N = 0; // Number of times to call recursive function           
  L = atol( argv[ 1 ] );
  N = atoi( argv[ 2 ] );
  Vector2f vPoint;
  vPoint.x = 0;
  vPoint.y = 0;
  // Create and Display Window
  PTree tree( L, N );
  return 0;
}  

(ptree.hpp)

#ifndef PTREE_H
#define PTREE_H
using namespace std;
using namespace sf;
class PTree /*:public sf::Drawable, public sf::Transformable*/{
public:
  // Constructor
  PTree( double L, int N );
  // Destructor
  ~PTree();
  // Recursive function to draw Pythagorias Tree
  void pTree( double L, int N, Vector2f vPoint, Vector2f vOrigin, float rotation );
private:
  float width = 0;
  float height = 0;
  int originX = 0;
  int originY = 0;
  float rotation = 0;
  RenderWindow window;
  int angle1 = 0;
  int angle2 = 0;
};
#endif // PTREE_H included

(ptree.cpp)

#include <SFML/System.hpp>
#include <SFML/Graphics.hpp>
#include <SFML/Window.hpp>
#include <math.h>
#include "PTree.hpp"
#include <iostream>
using namespace std;
using namespace sf;
// Constructor
PTree::PTree( double L, int N )
{ 
  width = ( 6 * L );
  height = ( 4 * L );
  Vector2f vPoint = { width/2, height - 1 };
  Vector2f vOrigin;
  vOrigin.x = L/2;
  vOrigin.y = L;
  /* vPoint.x = width/2;
  vPoint.y = height - 1;
 */
  window.create( VideoMode( width, height ), "Pythagoras Fractal Tree" );
  pTree( L, N, vPoint, vOrigin, 0 );
}
// Destructor
PTree::~PTree(){}
/*###########################################################################*/
// Recursive function to draw Pythagorias Tree
void PTree::pTree( double L, int N, Vector2f vPoint, Vector2f vOrigin, float rotation  )
{
  Vector2f vPointR;
  if( N < 1 )
  {
    return;
  }
  // Define a convex shape called convexSquare
  ConvexShape convexSquare( 4 );
  convexSquare.setPoint( 0, Vector2f( 0, 0 ));
  convexSquare.setPoint( 1, Vector2f( 0, L ));
  convexSquare.setPoint( 2, Vector2f( L, L ));
  convexSquare.setPoint( 3, Vector2f( L, 0 ));
  convexSquare.setOutlineThickness( 1.f );
  convexSquare.setFillColor( Color::Black );
  convexSquare.setOutlineColor( Color::White );
  convexSquare.setPosition( vPoint );
  convexSquare.setOrigin( vOrigin );
  convexSquare.setRotation( rotation );
  while( window.isOpen( ))
  {
    Event event;
    while( window.pollEvent( event ))
    {
      if( event.type == Event::Closed )
      {
        window.close( );
      }
    }
    if( N >= 0 )
    {
    window.draw( convexSquare );
    window.display( );
    L = ( L * ( sqrt(2)/2 ));
    N = N - 1;
    rotation = rotation - 135;
    cout << "LOOPS:" << N << endl;
//left
    vPoint = convexSquare.getTransform( ).transformPoint(   convexSquare.getPoint( 0 ));
    vOrigin = convexSquare.getPoint( (angle1) );
    pTree( L, N, vPoint, vOrigin, rotation );
    angle1 = (( angle1 + 1 ) % 4 );
//right
    vPointR = convexSquare.getTransform( ).transformPoint( convexSquare.getPoint( 3 ));
    vOrigin = convexSquare.getPoint( 2 );
    pTree( L, N, vPointR, vOrigin, rotation-90 ); 
    }
  }
cout << "X value =" << vPoint.x << "     Y value = " << vPoint.y << endl;

到目前为止,我试图将凸形的各个点返回到函数ptree的第二个递归调用中。这也没有显示任何内容。

最初,我仅在每个递归调用之前都使用vector2f vpoint并对其进行修改,但是在解决方案的知识库之后,我为右侧正方形创建了一个新变量,专门针对称为vector2f vpointr。

SFML文档没有为像我这样的菜鸟提供足够的例子。API本质上是一个选项列表,如果每个功能都有最小的示例。我以最好的能力搜索了互联网,以查看我是否通过错误的观点但找不到答案。

确实有效的一件事(尽管不完全正确)是当我切换递归电话...这意味着我在呼叫左侧正方形之前将右侧正方形的呼叫移动了左侧Quares没有显示。

在这一点

我试图递归显示这些正方形的方式是否存在问题?

我不确定除了堆栈溢出以寻求帮助外,我不确定从这里去哪里。

感谢您的时间和专业知识。

不要递归地调用整个while循环。仅临时调用图纸部分

// Initialize window...
while (window.isOpen())
{
    sf::Event event;
    // Handle events...
    window.clear();
    // call the recursive function here
    window.display();
}

您也可能要使用sf::RectangleShape绘制而不是sf::ConvexShape

这是一个工作的"示例":

#include <SFML/Graphics.hpp>
#include <cmath>
void drawPythagoreanTree(sf::RenderTarget&, const float, const int);
int main()
{
    const float L = 150;
    const int N = 14;
    const unsigned width = static_cast<unsigned>(6 * L);
    const unsigned height = static_cast<unsigned>(4 * L);
    sf::RenderWindow window{{width, height}, "Pythagorean Tree"};
    while (window.isOpen())
    {
        for (sf::Event event; window.pollEvent(event);)
        {
            if (event.type == sf::Event::Closed)
                window.close();
        }
        window.clear(sf::Color::White);
        drawPythagoreanTree(window, L, N);
        window.display();
    }
}
void drawPythagoreanTree(sf::RenderTarget& target, const int N,
                         const sf::RectangleShape& parent)
{
    static const float halfSqrt2 = sqrt(2.f) / 2;
    if (N < 1) return;
    target.draw(parent);
    auto const& sz = parent.getSize();
    auto const& tf = parent.getTransform();
    auto childL = parent;                    // copy parent's color and rotation
    childL.setSize(sz * halfSqrt2);          // resize
    childL.setOrigin(0, childL.getSize().y); // bottom left corner
    childL.setPosition(tf.transformPoint({0, 0})); // reposition
    childL.rotate(-45);
    drawPythagoreanTree(target, N - 1, childL);
    auto childR = parent;               // copy parent's color and rotation
    childR.setSize(sz * halfSqrt2);     // resize
    childR.setOrigin(childR.getSize()); // bottom right corner
    childR.setPosition(tf.transformPoint({sz.x, 0})); // reposition
    childR.rotate(45);
    drawPythagoreanTree(target, N - 1, childR);
}
void drawPythagoreanTree(sf::RenderTarget& target, const float L, const int N)
{
    sf::RectangleShape rect{{L, L}};
    // set origin to center of the rect, easier to center position on screen
    rect.setOrigin(rect.getSize() / 2.f);
    rect.setPosition(target.getSize().x / 2.f, target.getSize().y - L / 2.f);
    rect.setFillColor(sf::Color::Black);
    drawPythagoreanTree(target, N, rect);
}