How to use condition to check if typename T is integer type of float type in C++

Question

I am going to write a template to generate a vector of random data. The problem is std::uniform_int_distribution only accepts integer type, and std::uniform_real_distribution for float type. I want to combine both. Here is my code.

#include <vector>
#include <random>
#include <algorithm>
#include <iterator>
#include <functional>

template<typename T>
std::vector<T> generate_vector(size_t N, T lower = T(0), T higher = T(99)) {
    // Specify the engine and distribution. 
    if constexpr (std::is_integral<T>) {
    std::uniform_int_distribution<T> distribution(lower, higher);
    }
    else if constexpr (std::is_floating_point<T>) {
    std::uniform_real_distribution<T> distribution(lower, higher);
    }
    std::mt19937 engine; // Mersenne twister MT19937
    auto generator = std::bind(distribution, engine);
    std::vector<T> vec(N);
    std::generate(vec.begin(), vec.end(), generator);
    return vec;

I am confusing how to implement statements within if conditions. Integer type should include:short, int, long, long long, unsigned short, unsigned int, unsigned long, or unsigned long long. Float type includes float, double, or long double.

Any help suggestion?

Answer 1

As Justin points out in his comment it is simple enough to use an if constexpr block in the following way:

#include <type_traits>

if constexpr (std::is_integral_v<T>) {  // constexpr only necessary on first statement
    ...
} else if (std::is_floating_point_v<T>) {  // automatically constexpr
    ...
}

This is only available C++17. See the C++ references for more information on compile-time type information:

if constexpr (since C++17)

<type_traits> (since C++11)

constexpr specifier (since C++11)

Constant Expressions in general.