mirror of
https://github.com/Light-City/CPlusPlusThings.git
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dfe1c933b3
- 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
148 lines
3.3 KiB
Markdown
148 lines
3.3 KiB
Markdown
> 来自 [C++那些事网站](https://light-city.github.io/stories_things/cpp-guides/print-1-to-100-without-loop/) 的文章，由站点同步脚本导出。
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# C++ 无循环打印 1-100
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给大家设定一个条件：不使用循环，请打印出1-100个数字。
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如果使用循环很简单了，那么你能想到多少种方法呢？
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今天来给大家用一些C++的骚操作来写出这种代码，欢迎留言与转发～
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这道题目其实非常简单，但却涵盖了很多面试考点，其中包括：
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- 模版特化、便特化，如何编写递归模版
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- CRTP
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- std::copy、std::iota、std::transform如何使用，各自算法实现
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- 类静态变量
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- 等等
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## 1.递归
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1-100可以拆为`5*5*4`，用三个函数去打印它，函数a cout一次，然后变量加1，函数b层层嵌套a重复5次，函数c层层嵌套b重复5次，最后在main函数中对c层层嵌套4次。
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```cpp
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#include <iostream>
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int a(int n) {
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std::cout << n++ << std::endl;
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return n;
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}
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int b(int n) { return a(a(a(a(a(n))))); }
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int c(int n) { return b(b(b(b(b(n))))); }
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int main() {
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c(c(c(c(1))));
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return 0;
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}
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```
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## 2.模版重载
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使用模版与特化来处理，当迭代100时只输出100，否则输出I，并且+1递归处理。
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```cpp
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#include <iostream>
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template <int I>
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struct JustPrint : public JustPrint<I + 1> {
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JustPrint() : JustPrint<I + 1>() { std::cout << I << " "; }
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};
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template <>
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struct JustPrint<100> {
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JustPrint<100>() { std::cout << 1 << " "; }
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};
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int main() { JustPrint<1> j; }
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```
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## 3.CRTP
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方式3基于方式2，将其改造为crtp编程风格，是不是有点看不懂了，嘿嘿。
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下面struct可以被替换为class，这里我用类来描述。其实就是两个类，第一个类Print为父类，打印每次的I，并调用子类，子类PrintDerived即成Print，并将自己传递给父类，子类是一个重载类，控制递归终止。
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```cpp
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#include <iostream>
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template <typename Derived, int I>
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struct Print {
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Print() {
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std::cout << I << " ";
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Derived::next();
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}
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};
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template <int I>
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struct PrintDerived : public Print<PrintDerived<I>, I> {
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static void next() { PrintDerived<I + 1>(); }
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};
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template <>
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struct PrintDerived<100> : public Print<PrintDerived<100>, 100> {
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static void next() {}
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};
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int main() {
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PrintDerived<1> j;
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return 0;
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}
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```
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## 4.类静态变量
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创建一个Test数组，元素为100个，静态变量每次累加。
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```cpp
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#include <iostream>
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using namespace std;
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class Test {
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static int i;
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public:
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Test() { cout << ++i << endl; }
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};
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int Test::i;
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int main() {
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Test a[100];
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return 0;
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}
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```
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## 5.使用算法库
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使用`std::for_each`简单粗暴。
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```cpp
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int main() {
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std::vector<int> l(100);
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std::iota(l.begin(), l.end(), 1);
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std::for_each(l.begin(), l.end(), [](int& n) { std::cout << n << std::endl; });
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return 0;
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}
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```
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## 6.C++20 ranges
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使用C++20的std::views，在std::copy时打印。
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```cpp
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#include <iostream>
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#include <iterator>
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#include <ranges>
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int main() {
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auto numbers = std::views::iota(1, 100 + 1);
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std::copy(numbers.begin(), numbers.end(), std::ostream_iterator<int>(std::cout, "\n"));
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}
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```
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这个方式继续改造，还有一种实现，使用std::transform。
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```cpp
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auto range = std::views::transform(std::ranges::iota_view{1, 101}, [](const auto& x) {
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std::cout << x << std::endl;
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return x;
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});
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std::vector<int> a(range.begin(), range.end());
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``` |