C语言学习笔记:用C画一只小猪佩奇

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  • 来源:跟我学网络

在这里我们用到带符号距离场(signed distance field,SDF)表示圆形:用这个方法表示形状,同时我们用ASCII字符 |/=\ 画出形状的外框,并填充内部,类似如下:SDF的梯度代表SDF变化最大的方向,可以用这个方向去决定用哪一个字符。我们通过差分求SDF的梯度近似值,然后用atan2()求出梯度的角度:用C语言实现,在[-1,1]乘[-1,1]画布中画一个半径0.8并带有0.1宽度的外框的圆形:

#include <math.h>
#include <stdio.h> 
#define T double 
 T f(T x, T y)
 {
	  return sqrt(x * x + y * y) - 0.8f;
} 
char outline(T x, T y) 
{
	T delta = 0.001;
	if (fabs(f(x, y)) < 0.05) 
		{ 
			T dx = f(x + delta, y) - f(x - delta, y);
			T dy = f(x, y + delta) - f(x, y - delta);
			return "|/=\\|/=\\|"[(int)((atan2(dy, dx) / 6.2831853072 + 0.5) * 8 + 0.5)];
		} 
	else if (f(x, y) < 0)
		return '.'; 
	else 
		return ' '; 
}
 int main() 
{ 
 for (T y = -1; y < 1; y += 0.05, putchar('\n')) 
	 for (T x = -1; x < 1; x += 0.025) 
		 putchar(outline(x, y));
}

然后我们就可以画多个圆形,把它们适当的旋转或者缩放,再组合起来,就可以画出来小猪佩奇了:

// ASCII Peppa Pig by 凯撒ee
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#define T double
T c(T x,T y,T r)
{
	return sqrt(x*x+y*y)-r;
}
T u(T x,T y,T t)
{
	return x*cos(t)+y*sin(t);
}
T v(T x,T y,T t)
{
	return y*cos(t)-x*sin(t);
}
T fa(T x,T y)
{
	return fmin(c(x,y,0.5),c(x*0.47+0.15,y+0.25,0.3));
}
T no(T x,T y)
{
	return c(x*1.2+0.97,y+0.25,0.2);
}
T nh(T x,T y)
{
	return fmin(c(x+0.9,y+0.25,0.03),c(x+0.75,y+0.25,0.03));
}
T ea(T x,T y)
{
	return fmin(c(x*1.7+0.3,y+0.7,0.15),c(u(x,y,0.25)*1.7,v(x,y,0.25)+0.65,0.15));
}
T ey(T x,T y)
{
	return fmin(c(x+0.4,y+0.35,0.1),c(x+0.15,y+0.35,0.1));
}
T pu(T x,T y)
{
	return fmin(c(x+0.38,y+0.33,0.03),c(x+0.13,y+0.33,0.03));
}
T fr(T x,T y)
{
	return c(x*1.1-0.3,y+0.1,0.15);
}
T mo(T x,T y)
{
	return fmax(c(x+0.15,y-0.05,0.2),-c(x+0.15,y,0.25));
}
T o(T x,T y,T(*f)
(T,T),T i)
{
	T r=f(x,y);return fabs(r)<0.02?(atan2(f(x,y+1e-3)-r,f(x+1e-3,y)-r)+0.3)*1.273+6.5:r<0?i:0;
}
T s(T x,T y,T(*f)(T,T),T i)
{
	return f(x,y)<0?i:0;
}
T f(T x,T y)
{
	return o(x,y,no,1)?fmax(o(x,y,no,1),s(x,y,nh,12)):fmax(o(x,y,fa,1),fmax(o(x,y,ey,11),fmax(o(x,y,ea,1),fmax(o(x,y,mo,1),fmax(s(x,y,fr,13),s(x,y,pu,12))))));
}
int main(int a,char**b)
{
	for(T y=-1,s=a>1?strtod(b[1],0):1;
		y<0.6;y+=0.05/s,putchar('\n'))
	for(T x=-1;x<0.6;x+=0.025/s)
		putchar(" .|/=\\|/=\\| @!"[(int)f(u(x,y,0.3),v(x,y,0.3))]);
}

二倍:

四倍:八倍:一只不怎么精致的小猪猪就完成了!