Why can a std::tuple not be assigned with an initializer list?

Question

I wonder why this choice is made. It would allow to write many functions in a very clear and neat way.. for instance:

int greatestCommonDivisor(int a, int b)
{
    if (b > a)
        std::tie(a, b) = { b, a };
    while (b > 0)
        std::tie(a, b) = { b, a % b };
    return a;
}

Answer 1

std::initializer_list is a homogeneous collection of items, while std::tuple is heterogeneous. The only case where it makes sense to define a std::tuple::operator= for std::initializer_list is when the tuple is homogeneous and has the same size as the initializer list, which is a rare occurrence.

(Additional information in this question.)

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Solution/workaround: you can use std::make_tuple instead:

int greatestCommonDivisor(int a, int b)
{
    if (b > a)
        std::tie(a, b) = std::make_tuple(b, a);
    while (b > 0)
        std::tie(a, b) = std::make_tuple(b, a % b);
    return a;
}

...or std::tuple's constructor in C++17 (thanks to Template argument deduction for class templates):

int greatestCommonDivisor(int a, int b)
{
    if (b > a)
        std::tie(a, b) = std::tuple{b, a};
    while (b > 0)
        std::tie(a, b) = std::tuple{b, a % b};
    return a;
}