Share how to use C language to draw a piggy page using signed distance field (SDF) to represent a circle: file:///C:\Users\Administrator.WIN-
STED6B9V5UI\AppData\Local\Temp\ksohtml9656\wps19.png continues to use this method to represent shapes, but this time we want to use ASCII characters |/=\ to draw the outer frame of the shape and fill the inside, similar to this: =====
//.....\\
||.......||
\\.....//
===== The gradient of SDF represents the direction in which SDF changes the most, and this direction can be used to determine which character to use. file:///C:\Users\Administrator.WIN-
STED6B9V5UI\AppData\Local\Temp\ksohtml9656\wps20.png We obtain the gradient approximation of SDF by difference, and then use atan2() to find the angle of the gradient: file:///C:\Users\Administrator.WIN-
STED6B9V5UI\AppData\Local\Temp\ksohtml9656\wps21.jpg is simply implemented in C language. In the canvas, draw a circle with a radius of 0.8 and an outer frame of 0.1: file:///C:\Users\Administrator. WIN-
\AppData\Local\Temp\ksohtml9656\wps22.jpg #include #include #define T doubleT 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));} The code can be moved left and right! ▲ file:///C:\Users\Administrator.WIN-STED6B9V5UI\AppData\Local\Temp\ksohtml9656\wps23.png Then, we can draw multiple circles,
Rotate and scale them appropriately, combine them with structural solid geometry, then you can draw Peppa Pig with 19 lines of code: the code can move left and right!
▼// ASCII Peppa Pig by Milo Yip#include #include #include #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,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))]);}file:///C:\Users\Administrator.WIN-STED6B9V5UI\AppData\Local\Temp\ksohtml9656\wps24.png 2倍：
file:///C:\Users\Administrator.WIN-STED6B9V5UI\AppData\Local\Temp\ksohtml9656\wps25.png 4倍：
file:///C:\Users\Administrator.WIN-STED6B9V5UI\AppData\Local\Temp\ksohtml9656\wps26.png 8倍：
file:///C:\Users\Administrator.WIN-STED6B9V5UI\AppData\Local\Temp\ksohtml9656\wps27.png how about it? Is it meeting now? You can also try to make this Peppa move
In addition, if you want to better improve your programming ability, learn C language and C++ programming! Overtaking in a curve, one step faster! I may be able to help you here~
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