Book Image

Modern C++: Efficient and Scalable Application Development

By : Richard Grimes, Marius Bancila
Book Image

Modern C++: Efficient and Scalable Application Development

By: Richard Grimes, Marius Bancila

Overview of this book

C++ is one of the most widely used programming languages. It is fast, flexible, and used to solve many programming problems. This Learning Path gives you an in-depth and hands-on experience of working with C++, using the latest recipes and understanding most recent developments. You will explore C++ programming constructs by learning about language structures, functions, and classes, which will help you identify the execution flow through code. You will also understand the importance of the C++ standard library as well as memory allocation for writing better and faster programs. Modern C++: Efficient and Scalable Application Development deals with the challenges faced with advanced C++ programming. You will work through advanced topics such as multithreading, networking, concurrency, lambda expressions, and many more recipes. By the end of this Learning Path, you will have all the skills to become a master C++ programmer. This Learning Path includes content from the following Packt products: • Beginning C++ Programming by Richard Grimes • Modern C++ Programming Cookbook by Marius Bancila • The Modern C++ Challenge by Marius Bancila
Table of Contents (24 chapters)
Title Page
Copyright
About Packt
Contributors
Preface
12
Math Problems
13
Language Features
14
Strings and Regular Expressions
15
Streams and Filesystems
16
Date and Time
17
Algorithms and Data Structures
Index

Strings and numbers


The Standard Library contains various functions and classes to convert between C++ strings and numeric values.

Converting strings to numbers

The C++ standard library contains functions with names like stod and stoi that convert a C++ string object to a numeric value (stod converts to a double and stoi converts to an integer). For example:

    double d = stod("10.5"); 
    d *= 4; 
    cout << d << "n"; // 42

This will initialize the floating-point variable d with a value of 10.5, which is then used in a calculation and the result is printed on the console. The input string may have characters that cannot be converted. If this is the case then the parsing of the string ends at that point. You can provide a pointer to a size_t variable, which will be initialized to the location of the first character that cannot be converted:

    string str = "49.5 red balloons"; 
    size_t idx = 0; 
    double d = stod(str, &idx); 
    d *= 2; 
    string rest = str.substr...