Test your skills (2005)

[Andy]

since when did u conform to the norm wtd?

[wtd]

Andy wrote:

since when did u conform to the norm wtd?

Since I started wanting my code to work right.

Let me just reduce it to this: any time you spend learning nonstandard C++ or java, or whatever is time you've lost and you'll never get back. It's time you've not only not spent learning something useful, but it's work you have to undo later on, meaning even less time to learn useful stuff.

[wtd]

The answer to this one is way overdue.

wtd wrote:

For 30 bits, just to shake things up.

I have the following class in Ruby:

code:

class Foo    def initialize(bar, baz)       @bar, @baz = bar, baz    end end

Now, @bar and @baz are private, as all instance variables are in Ruby, but I can write methods which let me access those variables.

With only two additional lines of code add the ability to read and write to @bar, and read @baz.

If you use more than two lines, there will be no bits.

code:

class Foo    attr_accessor :bar    attr_reader :baz    def initialize(bar, baz)       @bar, @baz = bar, baz    end end

wtd wrote:

Next question!

I have a Haskell function for multiplying an integer by two. Its type is:

code:

double :: Int -> Int

Without changing the name "double", write the shortest possible function definition. This is a 20 bit question.

We start with the obvious:

code:

double n = n * 2

But we realize that n is redundant, so we use a partial application of the * operator, which is really just a function itself.

code:

double=(*2)

And that, folks, is as short as it gets.

[wtd]

The following will not compile. Make the minimal necessary change such that it does compile, and PM me with your answer.

c++:

class A {    private:       const int& b;    public:       A(const int& init_b)       {          b = init_b;       } }; int main() {    int foo(42);    A bar(foo);    return 0; }

[wtd]

Sorry, rizzix, that's incorrect.

[rizzix]

yea i deleted the post.. did't read it correctly try adding an &

[wtd]

Nope. Nowhere do you add an & in the solution.

[rizzix]

then it has to be the const.. but.. well i'm not fimiliar with the &.. anyhow.. then isint int & b = &k; the way to go.. and not int &b = k;

[Martin]

Here's a hint: what if you wanted to do this?

c++:

#include <iostream> class A {    private:       const int& b;    public:       A(const int& init_b)       {          std::cout << b << std::endl;          b = init_b;       } }; int main() {    int foo(42);    A bar(foo);    return 0; }

[rizzix]

huh? howz that a hint.. i got it already.. gee.. the alias is defined const. pfft. i shouldn't.. point is.. the referencing is wrong.. now isin't it?

as for my second question.. constructors cannot initialise const values?

[Martin]

the const is fine the way it is.

[wtd]

In C++, in addition to regular old C style pointers there're also references. A reference is declareed like so:

code:

int& foo

Whereas a regular int would be declared as:

code:

int foo;

And a pointer to an int as:

code:

int * foo;

A reference to a constant int would be like so:

code:

const int& foo;

If we have an int variable, like so:

code:

int foo(42);

We can assign it to an int reference like so:

code:

int& bar(foo);

The & used as a prefix operator returns the address in memory of the variable. This is useful when dealing with pointers, but not so much references.

For instance, we might have:

code:

int foo(42); int * bar(&foo);

[rizzix]

martin:i dont think so. i compiled it.. and apparently contstuctors cannot initialise const values.

[Martin]

Yes they can. I already got this one.

[wtd]

rizzix wrote:

martin:i dont think so. i compiled it.. and apparently contstuctors cannot initialise const values.

Oh, I assure you they can. Are you sure you understand C++ constructors as well as you think? What you think is initialization might not be initialization at all.