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When one interface inherits another, it is possible to declare a member in the derived interface that hides one defined by the base interface This happens when a member in a derived interface has the same declaration as one in the base interface In this case, the base interface name is hidden This will cause a warning message unless you specify the derived interface member with new

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Jun 21, 2015 · Exodus Conference 3- King Jesus (Word Aflame Ministries Whittier, Ca.) - Duration: 3:56 ...Duration: 3:49 Posted: Jun 21, 2015
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Here are Brad s comments about his rst draft: I m embarrassed to have you read it It s terrible! But having said that, I can now say that it s also pretty good By that I mean it provides some meat-and-potatoes content for me to work with What was your draft like Brad s was just about the right length 280 words, or about one page He made many grammatical errors, but he also made several excellent points How did you do

When implementing a member of an interface, it is possible to fully qualify its name with its interface name Doing this creates an explicit interface member implementation, or explicit implementation, for short For example, given

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At the moment of the problem, dh/dt = 2, r = 3, h = 10/3 Hence 0= 3 2+ 2 or 0 = 6 + (20 /3) dr dt dr 10 dt 3

interface IMyIF { int MyMeth(int x); }

then it is legal to implement IMyIF as shown here:

class MyClass : IMyIF { int IMyIFMyMeth(int x) { return x / 3; } }

As you can see, when the MyMeth( ) member of IMyIF is implemented, its complete name, including its interface name, is specified There are two reasons that you might need to create an explicit implementation of an interface method First, when you implement an interface method using its fully qualified name, you are providing an implementation that cannot be accessed through an object of the class Instead, it must be accessed via an interface reference Thus, an explicit implementation gives you a way to implement an interface method so that it is not a public member of the

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6 Of course 5000 = V = r 2 h We conclude that h= We wish to minimize A = (area of top) + (area of sides) = r 2 + 2 r h = r 2 + 2 r 5000 r 2 5000 r2

Part I:

class that provides the implementation Second, it is possible for a class to implement two interfaces, both of which declare methods by the same name and type signature Qualifying the names with their interfaces removes the ambiguity from this situation Let s look at an example of each The following program contains an interface called IEven, which defines two methods, IsEven( ) and IsOdd( ), which determine if a number is even or odd MyClass then implements IEven When it does so, it implements IsOdd( ) explicitly

// Explicitly implement an interface member using System; interface IEven { bool IsOdd(int x); bool IsEven(int x); } class MyClass : IEven { // Explicit implementation Notice that this member is private // by default bool IEvenIsOdd(int x) { if((x%2) != 0) return true; else return false; } // Normal implementation public bool IsEven(int x) { IEven o = this; // Interface reference to the invoking object return !oIsOdd(x); } } class Demo { static void Main() { MyClass ob = new MyClass(); bool result; result = obIsEven(4); if(result) ConsoleWriteLine("4 is even"); // result = obIsOdd(4); // Error, IsOdd not exposed // But, this is OK It creates an IEven reference to a MyClass object // and then calls IsOdd() through that reference IEven iRef = (IEven) ob; result = iRefIsOdd(3); if(result) ConsoleWriteLine("3 is odd"); } }

Thus the function to minimize is A(r ) = r 2 + Thus 0 = A (r ) = 2 r We find therefore that r3 = 5000 10000 r2 10000 r

12:

In the revision exercises that follow, use both Brad s draft and your own to give yourself two opportunities to apply the tools Whether your rst

Here is a program that illustrates this conversion operator:

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