Is this the best way to do a "with" statement in C++?

Question

Edit:

So this question was misinterpreted to such a ludicrous degree that it has no point anymore. I don't know how, since the question that I actually asked was whether my specific implementation of this—yes, known to be pointless, yes, not remotely resembling idiomatic C++—macro was as good as it could be, and whether it necessarily had to use auto, or if there was a suitable workaround instead. It was not supposed to generate this much attention, and certainly not a misunderstanding of this magnitude. It's pointless to ask respondents to edit their answers, I don't want anybody to lose reputation over this, and there's some good information floating around in here for potential future viewers, so I'm going to arbitrarily pick one of the lower-voted answers to evenly distribute the reputation involved. Move along, nothing to see here.

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I saw this question and decided it might be fun to write a with statement in C++. The auto keyword makes this really easy, but is there a better way to do it, perhaps without using auto? I've elided certain bits of the code for brevity.

template<class T>
struct with_helper {

    with_helper(T& v) : value(v), alive(true) {}

    T* operator->() { return &value; }
    T& operator*() { return value; }

    T& value;
    bool alive;

};


template<class T> struct with_helper<const T> { ... };


template<class T> with_helper<T>       make_with_helper(T& value) { ... }
template<class T> with_helper<const T> make_with_helper(const T& value) { ... }


#define with(value) \
for (auto o = make_with_helper(value); o.alive; o.alive = false)

Here's an (updated) usage example with a more typical case that shows the use of with as it is found in other languages.

int main(int argc, char** argv) {

    Object object;

    with (object) {

        o->member = 0;
        o->method(1);
        o->method(2);
        o->method(3);

    }

    with (object.get_property("foo").perform_task(1, 2, 3).result()) {

        std::cout
            << (*o)[0] << '\n'
            << (*o)[1] << '\n'
            << (*o)[2] << '\n';

    }

    return 0;

}

I chose o because it's an uncommon identifier, and its form gives the impression of a "generic thing". If you've got an idea for a better identifier or a more usable syntax altogether, then please do suggest it.

Answer 1

If you use auto, why use macros at all?

int main()
{
    std::vector<int> vector_with_uncommonly_long_identifier;

    {
        auto& o = vector_with_uncommonly_long_identifier;

        o.push_back(1);
        o.push_back(2);
        o.push_back(3);
    }

    const std::vector<int> constant_duplicate_of_vector_with_uncommonly_long_identifier
        (vector_with_uncommonly_long_identifier);

    {
        const auto& o = constant_duplicate_of_vector_with_uncommonly_long_identifier;

        std::cout
            << o[0] << '\n'
            << o[1] << '\n'
            << o[2] << '\n';
    }

    {
        auto o = constant_duplicate_of_vector_with_uncommonly_long_identifier.size();
        std::cout << o <<'\n';
    }
}

EDIT: Without auto, just use typedef and references.

int main()
{
    typedef std::vector<int> Vec;

    Vec vector_with_uncommonly_long_identifier;

    {
        Vec& o = vector_with_uncommonly_long_identifier;

        o.push_back(1);
        o.push_back(2);
        o.push_back(3);
    }
}

Answer 2

?? attempted vb syntax into C++

with says do all the things in the following block by default referencing the object I've said to do it with right? Executes a series of statements making repeated reference to a single object or structure.

with(a)
 .do
 .domore
 .doitall

so how is the example giving you the same syntax?

to me examples of why to use a with where multiple de referencess

so rather than

book.sheet.table.col(a).row(2).setColour
book.sheet.table.col(a).row(2).setFont
book.sheet.table.col(a).row(2).setText
book.sheet.table.col(a).row(2).setBorder

you have

with( book.sheet.table.col(a).row(2) )
  .setColour
  .setFont
  .setText
  .setBorder

seems just as easy, and more common syntax in C++ to

cell& c = book.sheet.table.col(a).row(2);
c.setColour
c.setFont
c.setText
c.setBorder

Answer 3

For C++0x (which you're assuming):

int main() {

    std::vector<int> vector_with_uncommonly_long_identifier;

    {
        auto& o = vector_with_uncommonly_long_identifier;

        o.push_back(1);
        o.push_back(2);
        o.push_back(3);

    }
}

Answer 4

Why not just use a good lambda?

auto func = [&](std::vector<int>& o) {
};
func(vector_with_a_truly_ridiculously_long_identifier);

The simple fact is that if your identifiers are so long, that you can't type them out every time, use a reference, function, pointer, etc to solve this problem, or better, refactor the name. Statements like this (e.g. using() in C#) have additional side effects (deterministic cleanup, in my example). Your statement in C++ has no notable actual benefits, since it doesn't actually invoke any additional behaviour against just writing the code out.