Book Image

C++20 STL Cookbook

By : Bill Weinman
Book Image

C++20 STL Cookbook

By: Bill Weinman

Overview of this book

Fast, efficient, and flexible, the C++ programming language has come a long way and is used in every area of the industry to solve many problems. The latest version C++20 will see programmers change the way they code as it brings a whole array of features enabling the quick deployment of applications. This book will get you up and running with using the STL in the best way possible. Beginning with new language features in C++20, this book will help you understand the language's mechanics and library features and offer insights into how they work. Unlike other books, the C++20 STL Cookbook takes an implementation-specific, problem-solution approach that will help you overcome hurdles quickly. You'll learn core STL concepts, such as containers, algorithms, utility classes, lambda expressions, iterators, and more, while working on real-world recipes. This book is a reference guide for using the C++ STL with its latest capabilities and exploring the cutting-edge features in functional programming and lambda expressions. By the end of the book C++20 book, you'll be able to leverage the latest C++ features and save time and effort while solving tasks elegantly using the STL.
Table of Contents (13 chapters)

Use if constexpr to simplify compile-time decisions

An if constexpr(condition) statement is used where code needs to be executed based on a compile-time condition. The condition may be any constexpr expression of type bool.

How to do it…

Consider the case where you have a template function that needs to operate differently depending upon the type of the template parameter.

template<typename T>
auto value_of(const T v) {
    if constexpr (std::is_pointer_v<T>) {
        return *v;  // dereference the pointer
    } else {
        return v;   // return the value
    }
}
int main() {
    int x{47};
    int* y{&x};
    cout << format("value is {}\n", value_of(x));  // value
    cout << format("value is {}\n", value_of(y));  
                                                // pointer
    return 0;
}

Output:

value is 47
value is 47

The type of the template parameter T is available at compile time. The constexpr if statement allows the code to easily distinguish between a pointer and a value.

How it works…

The constexpr if statement works like a normal if statement except it's evaluated at compile time. The runtime code will not contain any branch statements from a constexpr if statement. Consider our branch statement from above:

if constexpr (std::is_pointer_v<T>) {
    return *v;  // dereference the pointer
} else {
        return v;   // return the value
    }

The condition is_pointer_v<T> tests a template parameter, which is not available at runtime. The constexpr keyword tells the compiler that this if statement needs to evaluate at compile time, while the template parameter <T> is available.

This should make a lot of meta programming situations much easier. The if constexpr statement is available in C++17 and later.