docs: address community issue feedback (#5 #338 #142 #246 #136 #303)

- const README: rewrite the space-saving claim per issue #5 follow-ups
  (constant-expression const vs #define immediates; which const qualify
  as constant expressions)
- const README: fix the class-section diagnosis per issue #338 (overload
  resolution picks the const overload; the restriction only surfaces
  once that overload is removed)
- sizeof README: virtual-inheritance result is compiler-dependent
  (GCC merges vbptr+vfptr -> 16, MSVC keeps them separate -> 32) #246
- static README: new section on internal linkage / access restriction
  with anonymous-namespace comparison #142
- 10-day practice: full problem list table + fixed day links #136
- new design_pattern/singleton/singleton_default.hpp: reusable
  thread-safe CRTP singleton template, compile+concurrency tested #303

site_extras/cpp-projects-stack-calculator-design.md

@@ -0,0 +1,75 @@
+> 来自 [C++那些事网站](https://light-city.github.io/stories_things/cpp-projects/stack-calculator-design/) 的文章，由站点同步脚本导出。
+
+# C++ 栈式计算器设计
+
+写一个栈计算器，设计如下：
+
+- 支持 `+`、`-`、`*`、`/`运算
+- 支持后缀输入
+
+例如：
+```
+2 3 +
+```
+输出：
+```
+5
+```
+如何实现呢？
+
+其中最关键两个问题：
+
+1.如何分词
+
+2.如何实现栈
+
+## 1.如何分词
+
+首先定义好输入是一个字符串，空格风格，我们便可以使用istringstream分割每一个token了，基本框架为：
+
+```cpp
+std::istringstream iss(expression);
+std::string token;
+while (iss >> token) {
+}
+```
+
+## 2.如何实现栈
+
+对于计算器来说，最核心就是入栈、出栈。
+
+- 碰到数字入栈
+- 碰到操作符出栈计算
+
+我们便可以写出框架：
+
+```cpp
+if (isNumber(token)) {
+  pushNumber(token);
+} else if (isOperator(token)) {
+  popAndCal(token);
+} else {
+  throw std::runtime_error("Invalid token in expression.");
+}
+```
+
+popAndCal是计算的核心，如：
+
+```cpp
+void popAndCal(const std::string& token) {
+  if (operandStack.size() < 2) {
+    throw std::runtime_error("Invalid expression.");
+  } else {
+    double operand2 = operandStack.top();
+    operandStack.pop();
+    double operand1 = operandStack.top();
+    operandStack.pop();
+    double result = performOperation(operand1, operand2, token);
+    operandStack.push(result);
+  }
+}
+```
+
+那么我们在栈中变存储了我们想要的结果，最后返回top即可。
+
+是不是很简单，欢迎与我一起探讨。