为什么不是std :: iota constexpr？

以下程序打印出一张洗牌的牌组(整数):

#include 
#include 
#include 
#include 

typedef unsigned int card;
typedef std::array deck;
auto shuffled_deck(){
    deck d = {};
    std::iota(d.begin(), d.end(), 0);
    std::shuffle(d.begin(), d.end(), std::default_random_engine());
    return d;
}

int main(){
    for(auto& i: shuffled_deck()) std::cout << i << ", ";
}

编译g++ -std=c++17程序运行和打印:

18, 34, 27, 46, 11, 3, 12, 19, 33, 21, 41, 13, 36, 49, 40, 22, 8, 9, 28, 2, 6, 30, 50, 24, 37, 32, 35, 4, 15, 45, 47, 43, 14, 44, 20, 23, 29, 7, 31, 51, 26, 10, 42, 48, 0, 38, 5, 16, 17, 1, 25, 39,

这很好,但是直觉告诉我这个套牌可以在编译时创建,所以我制作了这个shuffled_deck方法constexpr

constexpr auto shuffled_deck(){
    deck d = {};
    std::iota(d.begin(), d.end(), 0); // Error! Iota isn't constexpr!
    std::shuffle(d.begin(), d.end(), std::default_random_engine());
    return d;
}

编译时g++ -std=c++17会给出编译错误,说明std::iota不是constexpr.我的问题是为什么？当然std::iota可以在编译时确定.标准库是否仅落后于此功能？