std::enable_if

Topic: std::enable_if

std::enable_if is a powerful template metaprogramming tool in C++ that allows you to control the instantiation of templates based on compile-time conditions. It's part of the C++ Standard Library and was introduced in C++11.

Basic Syntax

The basic syntax of std::enable_if is:

template <bool Condition, typename T = void>
struct enable_if;

If Condition is true, std::enable_if<Condition, T> defines a public member typedef type equal to T. If Condition is false, there is no member typedef.

Primary Use Cases

1. SFINAE (Substitution Failure Is Not An Error): Enable or disable function overloads based on type traits.
2. Template specialization: Control which template specialization is used based on type properties.
3. Compile-time assertions: Prevent instantiation of templates when certain conditions are not met.

Examples

Let's look at some examples to illustrate the use of std::enable_if:

1. Function Overloading

#include <type_traits>
#include <iostream>

// Function for integral types
template <typename T>
typename std::enable_if<std::is_integral<T>::value, bool>::type
is_odd(T i) {
    return bool(i % 2);
}

// Function for floating-point types
template <typename T>
typename std::enable_if<std::is_floating_point<T>::value, bool>::type
is_odd(T x) {
    return bool(int(x) % 2);
}

int main() {
    std::cout << "Is 5 odd? " << is_odd(5) << std::endl;
    std::cout << "Is 5.0 odd? " << is_odd(5.0) << std::endl;
}

In this example, we use std::enable_if to create two overloads of is_odd that work differently for integral and floating-point types.

2. Template Specialization

#include <type_traits>
#include <iostream>
#include <vector>

template <typename T, typename Enable = void>
class Container {
public:
    void add(const T& value) {
        std::cout << "General container" << std::endl;
    }
};

template <typename T>
class Container<T, typename std::enable_if<std::is_arithmetic<T>::value>::type> {
public:
    void add(const T& value) {
        std::cout << "Arithmetic container" << std::endl;
    }
};

int main() {
    Container<std::string> c1;
    c1.add("Hello");

    Container<int> c2;
    c2.add(42);
}

Here, we use std::enable_if to specialize the Container class for arithmetic types.

Advanced Usage: C++14 and Beyond

In C++14 and later, you can use std::enable_if_t as a shorthand:

template <typename T>
std::enable_if_t<std::is_integral<T>::value, bool>
is_even(T i) {
    return i % 2 == 0;
}

In C++17, you can use inline variables with constexpr if statements for cleaner syntax:

template <typename T>
auto is_even(T i) {
    if constexpr (std::is_integral_v<T>) {
        return i % 2 == 0;
    } else {
        static_assert(always_false<T>::value, "is_even only works for integral types");
    }
}

Considerations for Scientific Computing

In scientific computing and numerical analysis, std::enable_if can be particularly useful for:

1. Implementing different algorithms based on the precision of floating-point types.
2. Optimizing matrix operations for specific numeric types.
3. Ensuring correct template instantiations for mathematical functions.

For example, you might use it in a linear algebra library to select different algorithms based on the matrix type or size.

Conclusion

std::enable_if is a powerful tool in C++ template metaprogramming. It allows for more expressive and type-safe code, especially in library design and generic programming. As a professor in Scientific Computing, understanding and utilizing std::enable_if can lead to more efficient and flexible numerical algorithms and data structures in your C++ code.

Related

• How does std::enable_if improve code readability
• Can you provide examples of using std::enable_if with function templates
• What are the common pitfalls when using std::enable_if
• How does std::enable_if interact with SFINAE
• Are there any alternatives to std::enable_if in C++