vlambda
2020-01-17

惊呆了！C语言也能画小猪佩奇？【附源码】

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小猪佩奇那么火，那么我们接下来就分享下如何用 C 语言画小猪佩奇

使用带符号距离场（signed distance field, SDF）表示圆形：

沿用这个方法表示形状，但这次我们想利用 ASCII 字符|/=\画出形状的外框，并填充内部，类似这样：

   =====
 //.....\\
||.......||
 \\.....//
   =====

SDF 的梯度（gradient）代表 SDF 变化最大的方向，可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 语言简单实现，在画布中画一个半径 0.8 并带有 0.1 寛度外框的圆形：

#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));}

然后，我们就可以画多个圆形，把它们适当地旋转和缩放，用构造实体几何比它们组合起来，那么用 19 行代码就可以画出小猪佩奇了：

代码可以左右移动！▼

// 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[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))]);}