how to assign multiple values into a struct at once?

I can do this on initialization for a struct Foo:

Foo foo =  {bunch, of, things, initialized};

but, I can't do this:

Foo foo;
foo = {bunch, of, things, initialized};

So, two questions:

1. Why can't I do the latter, is the former a special constructor for initialization only?

2. How can I do something similar to the second example, i.e. declare a bunch of variables for a struct in a single line of code after it's already been initialized? I'm trying to avoid having to do this for large structs with many variables:

   Foo foo;
   
   foo.a = 1;
   foo.b = 2;
   foo.c = 3;
   //... ad infinitum
   

Try this:

Foo foo;
foo = (Foo){bunch, of, things, initialized};

This will work if you have a good compiler (e.g. GCC). You might need to turn on C99 mode with --std=gnu99, I'm not sure.

The first is an aggregate initializer - you can read up on those and tagged initializers at this solution:

What is tagged structure initialization syntax?

It is a special initialization syntax, and you can't do something similar after initialization of your struct. What you can do is provide a member (or non-member) function to take your series of values as parameters which you then assign within the member function - that would allow you to accomplish this after the structure is initialized in a way that is equally concise (after you've written the function the first time of course!)

In C++11 you can perform multiple assignment with "tie" (declared in the tuple header)

struct foo {
    int a, b, c;
} f;

std::tie(f.a, f.b, f.c) = std::make_tuple(1, 2, 3);

If your right hand expression is of fixed size and you only need to get some of the elements, you can use the ignore placeholder with tie

std::tie(std::ignore, f.b, std::ignore) = some_tuple; // only f.b modified

If you find the syntax std::tie(f.a, f.b, f.c) too code cluttering you could have a member function returning that tuple of references

struct foo {
    int a, b, c;
    auto members() -> decltype(std::tie(a, b, c)) {
        return std::tie(a, b, c);
    }
} f;

f.members() = std::make_tuple(1, 2, 3);

All this ofcourse assuming that overloading the assignment operator is not an option because your struct is not constructible by such sequence of values, in which case you could say

f = foo(1, 2, 3);