功能测试

Rratic

Zola

Markdown

封面图来自 Shadertoy: Planetary gears

你的系统支持 italicFontFace,并且你的浏览器(或者别的什么东西)继承了这一特性。

瞻彼阕者,虚室生白,吉祥止止。

功能评注:

  • `text` 会产生 text 的效果
    • 默认的 <code></code> 样式令人不悦。
      1. 使用 content: "" !important; 覆盖前后的反引号。
      2. 使用 text-decoration: 3px gold underline; 制作高亮。
    • Markdown 源文件中的链接无法自动转化同样令人不悦。
  • 这个列表的间距很好。
猫的类型颜色
橘猫#ffa9401
1

采自 Ant Design

#[derive(Reflect, Clone, Copy)]
#[reflect(SerializeWithRegistry, DeserializeWithRegistry)]
struct ComponentTypeLink(pub TypeId);

impl SerializeWithRegistry for ComponentTypeLink {
	fn serialize<S>(&self, serializer: S, registry: &TypeRegistry) -> Result<S::Ok, S::Error>
	where
		S: Serializer,
	{
		let registeration = registry.get(self.0).unwrap();
		let info = registeration.type_info();
		let path = info.type_path();
		serializer.serialize_str(path)
	}
}
- let mut me = self.entry::<FreeWill>.mut();
- world.execute(me);
+ if Some(mut me) = self.entry::<FreeWill>.get_mut() {
+     world.execute(me);
+ }
module Agda.Builtin.Bool where

data Bool : Set where
  false true : Bool

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Linkita

KaTeX

$\R^{1,3} \rtimes \operatorname{SO}(1,3)$ 是一个 $\set{A_n}$ 的 $\cancel{\boxed{~}}$.

$$ \begin{Vmatrix} a & b \cr c & d \end{Vmatrix} $$

$$ \begin{CD} A @>a>> B \cr @VbVV @AAcA \cr C @= D \end{CD} $$

Shortcodes

  graph LR;
	赤狐-->乙木;
	赤狐-->丙火;
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TODO

尚待完成。

That is not dead which can eternal lie,
And with strange aeons even death may die.

——H. P. Lovecraft

水曰润下,火曰炎上,木曰曲直,金曰从革,土爰稼穑。

——《尚书·周书·洪范第四》

What is a fish without an eye?

A fsh.

#Test

解析

**谓词(predicate)性质(property)**应该分别是粗体。

$12+34$

$$\left{x \in A \middle| |x| = 1 \right}$$

页面显示

$$Y f = (\lambda x. f(x x))(\lambda x. f(x x)) = (\lambda x. f(x x))(\lambda x. f(x x))(\lambda x. f(x x)) = f(Y f) = f((\lambda x. f(x x))(\lambda x. f(x x))) = f((\lambda x. f(x x))(\lambda x. f(x x))(\lambda x. f(x x))) = f(f(Y f))$$

abstract
摘要

$$Y f = (\lambda x. f(x x))(\lambda x. f(x x)) = (\lambda x. f(x x))(\lambda x. f(x x))(\lambda x. f(x x)) = f(Y f) = f((\lambda x. f(x x))(\lambda x. f(x x))) = f((\lambda x. f(x x))(\lambda x. f(x x))(\lambda x. f(x x))) = f(f(Y f))$$