alright. i'll stick with subtraction.

    On Wednesday, March 3, 2021, 10:59:38 AM EST, Gašper Ažman via Std-Proposals <std-proposals_at_[hidden]> wrote:
 
 Also to have an equivalent floating-point interface that can actually generate both 1.0 and 0.0

On Wed, Mar 3, 2021 at 3:02 PM Edward Catmur <ecatmur_at_[hidden]> wrote:

It seems to me that what you really want is a uniform_int_distribution that accepts a half-open interval. I assume the reason the Standard specifies a closed interval is to allow a distribution over the whole of an integer type?
On Wed, Mar 3, 2021 at 9:59 AM Gašper Ažman via Std-Proposals <std-proposals_at_[hidden]> wrote:

Not sure if I understand, because it seems so obvious otherwise: you just want a function that returns size()-1?

On Wed, Mar 3, 2021 at 7:56 AM Kurt Raburn via Std-Proposals <std-proposals_at_[hidden]> wrote:

hello,
i'm not sure if this was my fault, but this is the issue i was having.

std::uniform_int_distribution<int> dist(0, vec.size()); // runtime error

std::uniform_int_distribution<int> dist(0, vec.size() - 1); // ok

after some thought, i wrote a function that returns the index count.

template<typename T>
int indices(const std::vector<T>& vec)
{
    int cnt = 0;

    for (T i : vec) {
        ++cnt;
    }

    return --cnt;
}
then, i modified the size() function, and it returned the same result.

_NODISCARD size_type indices() const noexcept
        {
        return (static_cast<size_type>(this->_Mylast() - this->_Myfirst() - 1));
        }
usage:

std::uniform_int_distribution<int> dist(0, vec.indices());
forgive me if i overlooked a library function that would've solved this.

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