Use std::uniform_int_distribution and define its range later

Question

I have problem where I want to create a std::uniform_int_distribution in a struct and then give its range later. Below is what I want.

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Answer 1

You can just do

group.dis = std::uniform_int_distribution<>(0,19);

or

group.dis.param(std::uniform_int_distribution<>::param_type(0,19));

Another way would be to add a method to your struct

struct group_s {
    int k;
    std::uniform_int_distribution<> dis;
    void set(int a, int b) { dis = std::uniform_int_distribution<>(a,b); }
} group;


group.set(0,19);

Answer 2

You should do

group.dis = std::uniform_int_distribution<>(0,19);

instead of

group.dis(0,19);

Also, your code seems to be taken without reference directly from here, so a link as kind of a tributal citation would have been in order.

Answer 3

Use param():

using param_t = std::uniform_int_distribution<>::param_type;

group.dis.param(param_t(0, 19));

If the parameters change every time you use the distribution, then you can also consider using the two-parameter overload of operator() instead:

std::cout << group.dis(gen, param_t(0, 19)) << ' ';

As distribution objects are allowed to store extra bits of entropy obtained during a previous operator() call, this approach can be more efficient than constructing a new distribution object and assigning it.

Note that the cppreference page is incomplete and doesn't document the requirements the standard imposes on param_type. Given a distribution type D and its associated param_type P,

For each of the constructors of D taking arguments corresponding to parameters of the distribution, P shall have a corresponding constructor subject to the same requirements and taking arguments identical in number, type, and default values. Moreover, for each of the member functions of D that return values corresponding to parameters of the distribution, P shall have a corresponding member function with the identical name, type, and semantics.

(§26.5.1.6 [rand.req.dist]/p9)