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public char Pop() { if(tos==0) { ConsoleWriteLine(" -- Stack is empty"); return (char) 0; } tos--; return stck[tos]; } // Return true if the stack is full public bool IsFull() { return tos==stckLength; } // Return true if the stack is empty public bool IsEmpty() { return tos==0; } // Return total capacity of the stack public int Capacity() { return stckLength; } // Return number of objects currently on the stack public int GetNum() { return tos; } } // Demonstrate the Stack class class StackDemo { static void Main() { Stack stk1 = new Stack(10); char ch; int i; // Put some characters into stk1 ConsoleWriteLine("Push A through J onto stk1"); for(i=0; !stk1IsFull(); i++) stk1Push((char) ('A' + i)); // Create a copy of stck1 Stack stk2 = new Stack(stk1); // Display the contents of stk1 ConsoleWrite("Contents of stk1: "); while( !stk1IsEmpty() ) { ch = stk1Pop(); ConsoleWrite(ch); }

8:

ConsoleWriteLine(); ConsoleWrite("Contents of stk2: "); while ( !stk2IsEmpty() ) { ch = stk2Pop(); ConsoleWrite(ch); } ConsoleWriteLine("\n"); } }

The output is shown here:

In StackDemo, the first stack, stk1, is constructed and filled with characters This stack is then used to construct the second stack, stk2 This causes the following Stack constructor to be executed:

// Construct a Stack from a stack public Stack(Stack ob) { // Allocate memory for stack stck = new char[obstckLength]; // Copy elements to new stack for(int i=0; i < obtos; i++) stck[i] = obstck[i]; // Set tos for new stack tos = obtos; }

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Inside this constructor, an array is allocated that is long enough to hold the elements contained in the stack passed in ob Then, the contents of ob s array are copied to the new array, and tos is set appropriately After the constructor finishes, the new stack and the original stack are separate, but identical

Integration by parts is somewhat similar to the method of substitution in that the correct association will make a difficult integral into a not so difficult integral The fomula for integration by parts, which we will not derive or even justi@, is found in the Mathematical Tables Iudv = uv - Jvdu The key to successfid application of this rule is the correct initial choice of U and dv Sometimes you have to try more than one combination to get one to work well The purpose of the choice is to make the integral on the right side easier and not harder than the one you started with The best way to learn this is to go directly to some problems and see how it is done

When working with overloaded constructors, it is sometimes useful for one constructor to invoke another In C#, this is accomplished by using another form of the this keyword The general form is shown here: constructor-name(parameter-list1) : this(parameter-list2) { // body of constructor, which may be empty }

Part I:

Maryland Morgan State University (MArch) Institute of Architecture and Planning Baltimore, MD 21239 morganedu/academics/IAP/bsae_homehtml

When the constructor is executed, the overloaded constructor that matches the parameter list specified by parameter-list2 is first executed Then, if there are any statements inside the original constructor, they are executed Here is an example:

// Demonstrate invoking a constructor through this using System; class XYCoord { public int x, y; public XYCoord() : this(0, 0) { ConsoleWriteLine("Inside XYCoord()"); } public XYCoord(XYCoord obj) : this(objx, objy) { ConsoleWriteLine("Inside XYCoord(obj)"); } public XYCoord(int i, int j) { ConsoleWriteLine("Inside XYCoord(int, int)"); x = i; y = j; } } class OverloadConsDemo { static void Main() { XYCoord t1 = new XYCoord(); XYCoord t2 = new XYCoord(8, 9); XYCoord t3 = new XYCoord(t2); ConsoleWriteLine("t1x, t1y: " + t1x + ", " + t1y); ConsoleWriteLine("t2x, t2y: " + t2x + ", " + t2y); ConsoleWriteLine("t3x, t3y: " + t3x + ", " + t3y); } }

The output from the program is shown here:

Inside XYCoord(int, int) Inside XYCoord() Inside XYCoord(int, int) Inside XYCoord(int, int) Inside XYCoord(obj) t1x, t1y: 0, 0 t2x, t2y: 8, 9 t3x, t3y: 8, 9

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