Ivan Čukić KDAB CppCon 2021 INTRODUCTION DATA FUNCTIONS AND DATA ABSTRACTIONS FUNCTIONS COMPOSITION Doug McIlroy and Dennis Ritchie Ivan Čukić KDAB CppCon 2021 INTRODUCTION DATA FUNCTIONS AND DATA ABSTRACTIONS sed ${1}q Ivan Čukić KDAB CppCon 2021 INTRODUCTION DATA FUNCTIONS AND DATA ABSTRACTIONS FUNCTIONS Doug McIlroy, Bell System Technical Journal, 1978: Make each program do one thing well. To do a new job

consultant Authors of RealtimeSanitizerSpecial thanks Alistair Barker Co-author of RealtimeSanitizer Doug Wyatt Inventor/author of performance constraintsAgenda 1. Real-time programming 2. Existing strategies were designed to complement each otherUse both, and write better real-time code!Thanks! ● Doug Wyatt ● Ali Barker ● Reviews on LLVM work ○ @vitalybuka ○ @maskray ○ @sirride ○ @AaronBallman

pipeline in software engineering?  Pipes in Unix/Linux;  The famous competition: Donald Knuth vs. Doug McIlroy; 3. Overloading the pipe operator |  Overloaded pipe operator in std::ranges;

Visual Studio P2473 36RESOURCES Papers Modules in C++, 2004, Daveed Vandevoorde Modules, 2012, Doug Gregor A Module System for C++, 2014, Gabriel Dos Reis, Mark Hall, Gor Nishanov C++ Modules TS, 2018

2Prologue 3 ● Credentials ○ every library and application is open source ○ development using cutting edge C++ technology ○ source code hosted on github ○ prebuilt binaries are available on our download

threadsPrologue ● Credentials ○ every library and application is open source ○ projects are developed using cutting edge C++ technology ○ all source code hosted on github ○ prebuilt binaries available on our download

Frame Evaluation Requires Malloc Unbounded Memory UsagewHaT aBoUt tHe SaD pAtH pErFoRmAnCe!? 216Cutting C++ Exception time by +76% +88% ✨ By: Khalil EstellSome performance numbers Function Depth std::expected

，用取模运算得到余数 ? ，此时有： ? = 3? + ? 请注意，对于 ? ≤ 3 的边界情况，必须拆分出一个 1 ，乘积为 1 × (? − 1) 。 // === File: max_product_cutting.cpp === /* 最大切分乘积：贪心 */ int maxProductCutting(int n) { // 当 n <= 3 时，必须切分出一个 1 if (n <= 3) {

到余数 ? ，此时有： ? = 3? + ? 请注意，对于 ? ≤ 3 的边界情况，必须拆分出一个 1 ，乘积为 1 × (? − 1) 。 // === File: max_product_cutting.cpp === /* 最大切分乘积：贪心 */ int maxProductCutting(int n) { // 当 n <= 3 时，必须切分出一个 1 if (n <= 3) {

到余数 ? ，此时有： ? = 3? + ? 请注意，对于 ? ≤ 3 的边界情况，必须拆分出一个 1 ，乘积为 1 × (? − 1) 。 // === File: max_product_cutting.cpp === /* 最大切分乘积：贪心 */ int maxProductCutting(int n) { // 当 n <= 3 时，必须切分出一个 1 if (n <= 3) {