使用C ++繪制線條而沒有圖形
[英]Draw Line using c++ without graphics
問題描述
我希望編寫不使用圖形庫就使用c ++ lang繪制線的程序，我嘗試了很多，但沒有達到我想要的線方程為：
y=mx+b
我要畫這條線（例如）
y=3x+2
這是我的代碼
#include <iostream>
#include <math.h>
using namespace std;
int pth (int x) {
return 2*x-3;
}
int main() {
int i,j,y;
for (int x=-10;x <= 10;x++) {
for ( i=-10;i <= x;i++) {
if (i==x) {
y=pth(x);
cout<<"x=";
cout<<x;
cout<<"y=";
cout<<y;
for(j=-10;j<=y;j++) {
if(j==y)
cout << "*";
else
cout << " ";
}
}
}
cout << "\n";
}
cin.get();
return 0;
}
輸出是這樣的
x=0y=3 *
x=1y=5 *
x=2y=7 *
x=3y=9 *
x=4y=11 *
x=5y=13 *
x=6y=15 *
x=7y=17 *
x=8y=19 *
x=9y=21 *
x=10y=23 *
但是對於給定的方程（y = 2x-3），結果必須是
*
*
*
*
*
*
*
代碼是我認為正確的，但是問題是如何確定線條的方向，以便正確繪制
2 個解決方案
解決方案1
6 2017-12-11 17:25:33
因為我喜歡挑戰，所以我做了一個很小的Canvas類，其用法如下：
int main() {
using Canvas = BasicCanvas<160, 80>;
Canvas canvas;
canvas.origin = {canvas.cols()/3, canvas.rows()/3};
canvas.axes();
canvas.plot([](double x) { return x; });
canvas.plot([](double ) { return -8; });
canvas.plot([](double x) { return 3*log(x); });
canvas.plot([](double x) { return 4*sin(x/2); });
canvas.plot([](double x) { return 24*cos(x/12); });
std::cout << canvas;
}
哪些印刷品
或注釋掉原點分配：
履行
實現基本上遍歷x軸上的位置，並根據該點函數的角度（一階導數）繪制近似的線條圖字符：
template <size_t Columns = 100, size_t Rows = 50>
struct BasicCanvas {
using Line = std::array<char, Columns>;
using Screen = std::array<Line, Rows>;
struct Coord { size_t x, y; };
static constexpr size_t rows() { return Rows; }
static constexpr size_t cols() { return Columns; }
Screen screen;
Coord origin;
BasicCanvas(Coord origin = {Columns/2, Rows/2}) : origin(origin) {
Line empty;
std::fill(empty.begin(), empty.end(), '.');
std::fill(screen.begin(), screen.end(), empty);
}
friend std::ostream& operator<<(std::ostream& os, BasicCanvas const& c) {
for (auto& line : c.screen) {
os.write(line.data(), line.size()) << "\n";
}
return os;
}
Line& operator[](size_t y) { return screen.at(screen.size()-(y+1)); }
Line const& operator[](size_t y) const { return screen.at(screen.size()-(y+1)); }
char& operator[](Coord coord) { return operator[](coord.y).at(coord.x); }
char const& operator[](Coord coord) const { return operator[](coord.y).at(coord.x); }
void axes() {
for (auto& line : screen)
line.at(origin.x) = '|';
auto& y_axis = operator[](origin.y);
for (auto& cell : y_axis)
cell = '-';
y_axis.at(origin.x) = '+';
}
template <typename F>
void plot(F f, double scaleX = 1.0, double scaleY = 1.0) {
for (size_t x_tick = 0; x_tick < Columns; ++x_tick) {
auto x = (x_tick * scaleX) - origin.x;
auto y = f(x);
auto y_ = derivative(f, x, scaleX/2);
size_t y_tick = (y / scaleY) + origin.y;
if (y_tick < Rows)
operator[]({x_tick, y_tick}) = glyph(y_);
}
}
private:
template <typename F>
auto derivative(F const& f, double x, double dx = 0.01) {
return (f(x+dx)-f(x-dx))/(2*dx);
}
char glyph(double tangent) {
auto angle = atan(tangent);
while (angle < 0)
angle += 2*M_PI;
int angle_index = 2.0 * angle / atan(1);
return R"(--/||\--)"[angle_index % 8];
}
};
完整清單
（簡化功能選擇）：
#include <iostream>
#include <array>
#include <cmath>
template <size_t Columns = 100, size_t Rows = 50>
struct BasicCanvas {
using Line = std::array<char, Columns>;
using Screen = std::array<Line, Rows>;
struct Coord { size_t x, y; };
static constexpr size_t rows() { return Rows; }
static constexpr size_t cols() { return Columns; }
Screen screen;
Coord origin;
BasicCanvas(Coord origin = {Columns/2, Rows/2}) : origin(origin) {
Line empty;
std::fill(empty.begin(), empty.end(), ' ');
std::fill(screen.begin(), screen.end(), empty);
}
friend std::ostream& operator<<(std::ostream& os, BasicCanvas const& c) {
for (auto& line : c.screen) {
os.write(line.data(), line.size()) << "\n";
}
return os;
}
Line& operator[](size_t y) { return screen.at(screen.size()-(y+1)); }
Line const& operator[](size_t y) const { return screen.at(screen.size()-(y+1)); }
char& operator[](Coord coord) { return operator[](coord.y).at(coord.x); }
char const& operator[](Coord coord) const { return operator[](coord.y).at(coord.x); }
void axes() {
for (auto& line : screen)
line.at(origin.x) = '|';
auto& y_axis = operator[](origin.y);
for (auto& cell : y_axis)
cell = '-';
y_axis.at(origin.x) = '+';
}
template <typename F>
void plot(F f, double scaleX = 1.0, double scaleY = 1.0) {
for (size_t x_tick = 0; x_tick < Columns; ++x_tick) {
auto x = (x_tick * scaleX) - origin.x;
auto y = f(x);
auto y_ = derivative(f, x, scaleX/2);
size_t y_tick = (y / scaleY) + origin.y;
if (y_tick < Rows)
operator[]({x_tick, y_tick}) = glyph(y_);
}
}
private:
template <typename F>
auto derivative(F const& f, double x, double dx = 0.01) {
return (f(x+dx)-f(x-dx))/(2*dx);
}
char glyph(double tangent) {
auto angle = atan(tangent);
while (angle < 0)
angle += 2*M_PI;
int angle_index = 2.0 * angle / atan(1);
return R"(--/||\--)"[angle_index % 8];
}
};
int main() {
using Canvas = BasicCanvas<60, 30>;
Canvas canvas;
//canvas.origin = {canvas.cols()/3, canvas.rows()/3};
canvas.axes();
canvas.plot([](double x) { return x; });
//canvas.plot([](double ) { return -8; });
canvas.plot([](double x) { return 3*log(x); });
canvas.plot([](double x) { return 4*sin(x/2); });
//canvas.plot([](double x) { return 24*cos(x/12); });
std::cout << canvas;
}
打印
| /
| /
| /
| /
| / -
| / --------
| / ------
| / ----
| / ---
| /--
| --
--- --\ | /-- --\ /--
/ \ / | / \ /
/ \ |/ \ /
-----/-----\------------------/|----\------/----------------
/ \ /| \ /
/ \ //| \ /
\ / \ / | / \ /
--/ \-- --/ | \-- ---
/ |
/ |
/ |
/ |
/ |
/ |
/ |
/ |
/ |
/ |
/ |
解決方案2
0 2017-12-11 13:57:39
我看着你的“ C”問題。
如果要使用終端並在屏幕上放置字符，請為您的系統找到一個ansi終端仿真器，然后嘗試使用Ansi_t：
class Ansi_t // use ansi features of gnome-terminal
{
// note: Ubuntu 15.10 gnome-terminal ansi term cursor locations
// are 1-based with origin 1,1 at top left corner
enum ANSI : char { ESC = 27 };
public:
static inline std::string clrscr(void)
{
std::stringstream ss;
ss << static_cast<char>(ESC) << "[H" // home
<< static_cast<char>(ESC) << "[2J"; // clrbos
return(ss.str());
}
// r/c are 0 based------v------v------0 based
static inline std::string gotoRC(int r, int c)
{
std::stringstream ss;
// Note: row/col of GameOfLife_t is 0 based (as in C++)
ss << static_cast<char>(ESC) << "["
<< (r+1) << ';' << (c+1) << 'H';
// xy of ansi terminal is 1 based -------^^--------------^^
return(ss.str());
}
// tbr - add more ansi functions as needed
}; // Ansi_t
典型用法是“
std::cout << Ansi_t::clrscr() << std::flush; // clears the screen
std::cout << Ansi_t::gotoRC(5, 25) << '*' << std::flush;
// position the cursor, and output a single 'dot':
現在用它畫一條線，等等。
注-在ubuntu上，終端仿真通常具有熱鍵以減小字體大小-對於某些終端，此鍵為“ Ctrl減號”。 重復應用可將字體大小減小為不可讀，因此每個字符越來越像點。
我已經成功地將Ansi_t用於GameOfLife ...
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