C# Brainteasers - Answers

These are the answers for the C# brainteasers. I've included the questions again just for clarity.

1) Overloading

What is displayed, and why?

using System;

class Base
{
    public virtual void Foo(int x)
    {
        Console.WriteLine ("Base.Foo(int)");
    }
}

class Derived : Base
{
    public override void Foo(int x)
    {
        Console.WriteLine ("Derived.Foo(int)");
    }
    
    public void Foo(object o)
    {
        Console.WriteLine ("Derived.Foo(object)");
    }
}

class Test
{
    static void Main()
    {
        Derived d = new Derived();
        int i = 10;
        d.Foo(i);
    }
}

Answer: Derived.Foo(object) is printed - when choosing an overload, if there are any compatible methods declared in a derived class, all signatures declared in the base class are ignored - even if they're overridden in the same derived class!

2) Order! Order!

What will be displayed, why, and how confident are you?

using System;

class Foo
{
    static Foo()
    {
        Console.WriteLine ("Foo");
    }
}

class Bar
{
    static int i = Init();
    
    static int Init()
    {
        Console.WriteLine("Bar");
        return 0;
    }
}

class Test
{
    static void Main()
    {
        Foo f = new Foo();
        Bar b = new Bar();
    }
}

Answer: On my box, Bar is printed and then Foo. This is because Foo has a static constructor, which cannot be run until the exact point at which the class first has to be initialized. Bar doesn't have a static constructor though, so the CLR is allowed to initialize it earlier. However, there's nothing to guarantee that Bar will be printed at all. No static fields have been referenced, so in theory the CLR doesn't have to initialize it at all in our example. This is all due to the beforefieldinit flag.

3) Silly arithmetic

Computers are meant to be good at arithmetic, aren't they? Why does this print "False"?

double d1 = 1.000001;
double d2 = 0.000001;
Console.WriteLine((d1-d2)==1.0);

Answer: All the values here are stored as binary floating point. While 1.0 can be stored exactly, 1.000001 is actually stored as 1.0000009999999999177333620536956004798412322998046875, and 0.000001 is actually stored as 0.000000999999999999999954748111825886258685613938723690807819366455078125. The difference between them isn't exactly 1.0, and in fact the difference can't be stored exactly either. Learn more about binary floating point

4) Print, print, print...

Here's some code using the anonymous method feature of C# 2. What does it do?

using System;
using System.Collections.Generic;

class Test
{
    delegate void Printer();
    
    static void Main()
    {
        List<Printer> printers = new List<Printer>();
        for (int i=0; i < 10; i++)
        {
            printers.Add(delegate { Console.WriteLine(i); });
        }
        
        foreach (Printer printer in printers)
        {
            printer();
        }
    }
}

Answer: Ah, the joys of captured variables. There's only one i variable here, and its value changes on each iteration of the loop. The anonymous methods capture the variable itself rather than its value at the point of creation - so the result is 10 printed ten times!

5) Literally nothing wrong with the compiler here...

Should this code compile? Does it? What does it mean?

using System;

class Test
{
    enum Foo { Bar, Baz };
    
    static void Main()
    {
        Foo f = 0.0;
        Console.WriteLine(f);
    }
}

Answer: This shouldn't compile, but it does under the MS compilers for both C# 2 and 3 (and probably 1 as well - I haven't checked). It shouldn't compile because only the literal 0 should be implicitly convertible to the default value of any enum. Here the decimal is 0.0. Just a little compiler bug. The result is to print Bar as that's the 0 value of the Foo.

More along the same lines...

using System;

class Test
{
    enum Foo { Bar, Baz };
    
    const int One = 1;
    const int Une = 1;
    
    static void Main()
    {
        Foo f = One-Une;
        Console.WriteLine(f);
    }
}

Answer: Eek - it gets worse! This won't compile under the MS C# 2 compiler, but will compile with the MS C# 3 compiler. It's a known bug due to some optimisation being done too early, collecting constants of 0 and thinking that any known 0 constant should be convertible to the 0 value of any enum. It's with us now, and unlikely to ever be fixed as it could break some code which is technically illegal but working perfectly well. It's possible that the spec will change instead, of course.

6) Type inference a-go-go

I first saw this on Ayende's blog (in a rather more obscure form, admittedly). Once again, work out what will be printed, and why.

using System;

class Test
{
    static void Main()
    {
        Foo("Hello");
    }
    
    static void Foo(object x)
    {
        Console.WriteLine("object");
    }
    
    static void Foo<T>(params T[] x)
    {
        Console.WriteLine("params T[]");
    }
}

Answer: params T[] is printed. Now why would the compiler choose to create an array when it doesn't have to? Well... there are two stages to this. Firstly, when trying to find overloads which are legitimate candidates to be called, type inference works out that T should be System.String. Nothing scary so far.

Then overloading tries to work out which method is "better". If it's a choice between string x and params string[] x the former will always win - but at this point it's effectively a choice between object x and params string[] x. (The fact that one is actually a generic method is only relevant in a tie-break situation.) For the purposes of working out "better conversions" the expanded form of the method with the params parameter is then used. This means that by the time actual conversions are considered, the choices are object x or string x - so clearly the latter wins.



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