std::if_enable

Class: std::enable_if

std::enable_if is a powerful template metaprogramming tool in C++ that allows for conditional compilation of function templates based on compile-time conditions. It is commonly used for SFINAE (Substitution Failure Is Not An Error) techniques and to enable or disable function overloads based on type traits.

Example 1: Basic Usage

#include <iostream>
#include <type_traits>

template <typename T>
typename std::enable_if<std::is_integral<T>::value, bool>::type
is_odd(T i) {
    return i % 2 != 0;
}

int main() {
    std::cout << "Is 5 odd? " << is_odd(5) << std::endl;
    // std::cout << "Is 3.14 odd? " << is_odd(3.14) << std::endl; // This would cause a compile error
    return 0;
}

Explanation

• The is_odd function template is only enabled for integral types.
• std::enable_if<std::is_integral<T>::value, bool>::type is used as the return type.
• If T is not an integral type, the function will not be part of the overload set, causing a compile-time error if called with a non-integral type.

Example 2: Function Overloading

#include <iostream>
#include <type_traits>
#include <string>

class MyClass {
public:
    template <typename T>
    typename std::enable_if<std::is_integral<T>::value, void>::type
    print(T value) {
        std::cout << "Integral type: " << value << std::endl;
    }

    template <typename T>
    typename std::enable_if<std::is_floating_point<T>::value, void>::type
    print(T value) {
        std::cout << "Floating-point type: " << value << std::endl;
    }

    template <typename T>
    typename std::enable_if<std::is_same<T, std::string>::value, void>::type
    print(const T& value) {
        std::cout << "String type: " << value << std::endl;
    }
};

int main() {
    MyClass obj;
    obj.print(42);
    obj.print(3.14);
    obj.print(std::string("Hello"));
    return 0;
}

Explanation

• This example demonstrates how std::enable_if can be used to create function overloads based on type traits.
• Three print function templates are defined, each enabled for a specific type category:
• Integral types
• Floating-point types
• String type
• The appropriate function is selected at compile-time based on the argument type.

Example 3: Template Specialization

#include <iostream>
#include <type_traits>

template <typename T, typename Enable = void>
struct TypeInfo {
    static constexpr const char* name = "Unknown";
};

template <typename T>
struct TypeInfo<T, typename std::enable_if<std::is_integral<T>::value>::type> {
    static constexpr const char* name = "Integer";
};

template <typename T>
struct TypeInfo<T, typename std::enable_if<std::is_floating_point<T>::value>::type> {
    static constexpr const char* name = "Floating-point";
};

int main() {
    std::cout << "int is: " << TypeInfo<int>::name << std::endl;
    std::cout << "double is: " << TypeInfo<double>::name << std::endl;
    std::cout << "char is: " << TypeInfo<char>::name << std::endl;
    std::cout << "std::string is: " << TypeInfo<std::string>::name << std::endl;
    return 0;
}

Explanation

• This example shows how std::enable_if can be used for template specialization.
• A primary template TypeInfo is defined with a default "Unknown" type name.
• Two specializations are created using std::enable_if:
• One for integral types
• One for floating-point types
• The appropriate specialization is selected based on the type traits of T.

Summary

std::enable_if is a versatile tool in C++ template metaprogramming that allows for:

1. Conditional compilation of function templates
2. SFINAE-based function overloading
3. Template specialization based on type traits

It helps in writing more expressive and type-safe code by enabling or disabling certain template instantiations based on compile-time conditions. This can lead to better error messages, improved code organization, and more efficient compile-time type checking.

When using std::enable_if, it's important to consider: - The readability of the code, as complex enable_if conditions can be hard to understand - The potential for increased compile times due to more complex template instantiations - The need for fallback implementations or appropriate error messages when no enabled overload is found

Overall, std::enable_if is a powerful feature that, when used judiciously, can greatly enhance the flexibility and robustness of template-based C++ code.

Citations:

[1] https://leimao.github.io/blog/CPP-Enable-If/ [2] https://stackoverflow.com/questions/6972368/stdenable-if-to-conditionally-compile-a-member-function [3] https://learn.microsoft.com/en-us/cpp/standard-library/enable-if-class?view=msvc-170 [4] https://cplusplus.com/reference/type_traits/enable_if/ [5] https://www.codeproject.com/Articles/878913/enable-if

Related

• How does std::enable_if differ from std::enable_if_t
• Can you provide a practical example of using std::enable_if in a real-world application
• What are some common pitfalls when using std::enable_if
• How does std::enable_if interact with SFINAE
• Are there any alternatives to std::enable_if for template specialization