I want to implement a insertion method for a Binary search tree, and come up with a solution below. I know there are plenty of code examples but I wonder what is the problem in my implementation? Or is there a problem? When I had traced it I thought I have missed something.
public void insertBST(Node<Double> head, int value){
if (head == null){
head = new Node<Double>(value);
return;
}
else {
if (head.getValue() > value)
insertBST(head.getLeft(), value);
else
insertBST(head.getRight(), value);
}
}
When you reassign a passed parameter, you're only changing the local variable, not the value passed to the function. You can read this question for more information - Is Java "pass-by-reference"? This is Java, right? Either way, a similar argument likely applies.
This is the problem with this line of code:
head = new Node<Double>(value);
You aren't changing the value passed into the function, so you never add to the tree.
You have two alternatives here, either the option presented by amdorra, or returning the current node:
public void insertBST(Node<Double> current, int value)
{
if (current == null)
{
return new Node<Double>(value);
}
else
{
if (head.getValue() > value)
head.setLeft(insertBST(head.getLeft(),value));
else
head.setRight(insertBST(head.getRight(),value));
return current;
}
}
To call the function, you can simply say:
root = insertBST(root, value);
With alternatives, the root will have to be handled as a special case.
at the beginning of you function you are adding the new Node to a part you will never have access to outside this function
so i will assume that your Node class looks like the following
Class Node{
private Node left;
private Node right;
//constructor, setters and getters and stuff
}
you could modify your code to look like the following:
if (head.getValue() > value){
if(head.getLeft == null) {
head.setLeft(new Node<Double>(value));
return;
}
insertBST(head.getLeft(),value);
}
else{
if(head.getRight == null) {
head.setRight(new Node<Double>(value));
return;
}
insertBST(head.getRight(),value);
}
you should also remove this part if (head==null) and always make sure you are sending a valid Node to the first call
hopefully you can help me. First of all, let me explain what my problem is.
I have two ViewModels. The first one has e.g. stored information in several textboxes.
For example
private static string _tbxCfgLogfile;
public string TbxCfgLogfile
{
get { return _tbxCfgLogfile; }
set
{
_tbxCfgLogfile = value;
NotifyOfPropertyChange(() => TbxCfgLogfile);
}
}
The other ViewModel has a Button where i want to save this data from the textboxes.
It does look like this
public bool CanBtnCfgSave
{
get
{
return (new PageConfigGeneralViewModel().TbxCfgLogfile.Length > 0 [...]);
}
}
public void BtnCfgSave()
{
new Functions.Config().SaveConfig();
}
How can i let "CanBtnCfgSave" know that the condition is met or not?
My first try was
private static string _tbxCfgLogfile;
public string TbxCfgLogfile
{
get { return _tbxCfgLogfile; }
set
{
_tbxCfgLogfile = value;
NotifyOfPropertyChange(() => TbxCfgLogfile);
NotifyOfPropertyChange(() => new ViewModels.OtherViewModel.CanBtnCfgSave);
}
}
It does not work. When i do remember right, i can get the data from each ViewModel, but i cannot set nor Notify them without any effort. Is that right? Do i have to use an "Event Aggregator" to accomplish my goal or is there an alternative easier way?
Not sure what you are doing in your viewmodels - why are you instantiating viewmodels in property accessors?
What is this line doing?
return (new PageConfigGeneralViewModel().TbxCfgLogfile.Length > 0 [...]);
I can't be sure from your setup as you haven't mentioned much about the architecture, but sincce you should have an instance of each viewmodel, there must be something conducting/managing the two (or one managing the other)
If you have one managing the other and you are implementing this via concrete references, you can just pick up the fields from the other viewmodel by accessing the properties directly, and hooking the PropertyChanged event of the child to notify the parent
class ParentViewModel : PropertyChangedBase
{
ChildViewModel childVM;
public ParentViewModel()
{
// Create child VM and hook up event...
childVM = new ChildViewModel();
childVM.PropertyChanged = ChildViewModel_PropertyChanged;
}
void ChildViewModel_PropertyChanged(object sender, System.ComponentModel.PropertyChangedEventArgs e)
{
// When any properties on the child VM change, update CanSave
NotifyOfPropertyChange(() => CanSave);
}
// Look at properties on the child VM
public bool CanSave { get { return childVM.SomeProperty != string.Empty; } }
public void Save() { // do stuff }
}
class ChildViewModel : PropertyChangedBase
{
private static string _someProperty;
public string SomeProperty
{
get { return _someProperty; }
set
{
_someProperty = value;
NotifyOfPropertyChange(() => SomeProperty);
}
}
}
Of course this is a very direct way to do it - you could just create a binding to CanSave on the child VM if that works, saving the need to create the CanSave property on the parent
In my windows phone app, I need to track some events to get a good flow. But I'm not sure how to handle them in good sequence.
What needs to be done at startup of the app:
Main view is loaded and corresponding view model instantiated
In the constructor of the view model I initiate a login sequence that signals when completed with an eventhandler
Now when the login sequence has finished AND the view is completely loaded I need to startup another sequence.
But here is the problem, the order of these 2 events 'completing' is not always the same...
I've use the EventToCommand from MVVMLight to signal the view model that the view has 'loaded'.
Any thoughts on how to synchronize this.
As you should not use wait handles or something similar on the UI thread. You will have to sync the two method using flags in your view model and check them before progressing.
So, implement two boolean properties in your view model. Now when the login dialog is finished set one of the properties (lets call it IsLoggedIn) to true, and when the initialization sequence is finished you set the other property (how about IsInitialized) to true. The trick now lies in the implementation of the setter of these two properties:
#region [IsInitialized]
public const string IsInitializedPropertyName = "IsInitialized";
private bool _isInitialized = false;
public bool IsInitialized {
get {
return _isInitialized;
}
set {
if (_isInitialized == value)
return;
var oldValue = _isInitialized;
_isInitialized = value;
RaisePropertyChanged(IsInitializedPropertyName);
InitializationComplete();
}
}
#endregion
#region [IsLoggedIn]
public const string IsLoggedInPropertyName = "IsLoggedIn";
private bool _isLoggedIn = false;
public bool IsLoggedIn {
get {
return _isLoggedIn;
}
set {
if (_isLoggedIn == value)
return;
var oldValue = _isLoggedIn;
_isLoggedIn = value;
RaisePropertyChanged(IsLoggedInPropertyName);
InitializationComplete();
}
}
#endregion
public void InitializationComplete() {
if (!(this.IsInitialized && this.IsLoggedIn))
return;
// put your code here
}
Alternatively you can remove the InitializationComplete from the setters and change InitializationComplete to:
public void InitializationComplete() {
// put your code here
}
Then subscribe to the 'PropertyChanged' event use the following implementation:
private void Class1_PropertyChanged(object sender, System.ComponentModel.PropertyChangedEventArgs e) {
if (e.PropertyName == IsInitializedPropertyName || e.PropertyName == IsLoggedInPropertyName) {
if (this.IsInitialized && this.IsLoggedIn)
InitializationComplete();
}
}
I have a series of QTextEdits and QLineEdits connected to a slot through a QSignalMapper(which emits a textChanged(QWidget*) signal). When the connected slot is called (pasted below), I need to be able to differentiate between the two so I know whether to call the text() or toPlainText() function. What's the easiest way to determine the subclass type of a QWidget?
void MainWindow::changed(QWidget *sender)
{
QTextEdit *temp = qobject_cast<QTextEdit *>(sender);
QString currentText = temp->toPlainText(); // or temp->text() if its
// a QLineEdit...
if(currentText.compare(""))
{
...
}
else
{
...
}
}
I was considering using try-catch but Qt doesn't seem to have very extensive support for Exceptions... Any ideas?
Actually, your solution is already almost there. In fact, qobject_cast will return NULL if it can't perform the cast. So try it on one of the classes, if it's NULL, try it on the other:
QString text;
QTextEdit *textEdit = qobject_cast<QTextEdit*>(sender);
QLineEdit *lineEdit = qobject_cast<QLineEdit*>(sender);
if (textEdit) {
text = textEdit->toPlainText();
} else if (lineEdit) {
text = lineEdit->text();
} else {
// Return an error
}
You can also use sender->metaObject()->className() so you won't make unnecesary casts. Specially if you have a lot of classes to test. The code will be like this:
QString text;
QString senderClass = sender->metaObject()->className();
if (senderClass == "QTextEdit") {
QTextEdit *textEdit = qobject_cast<QTextEdit*>(sender);
text = textEdit->toPlainText();
} else if (senderClass == "QLineEdit") {
QLineEdit *lineEdit = qobject_cast<QLineEdit*>(sender);
text = lineEdit->text();
} else {
// Return an error
}
I know is an old question but I leave this answer just in case it would be useful for somebody...
I am getting some errors thrown in my code when I open a Windows Forms form in Visual Studio's designer. I would like to branch in my code and perform a different initialization if the form is being opened by designer than if it is being run for real.
How can I determine at run-time if the code is being executed as part of designer opening the form?
if (System.ComponentModel.LicenseManager.UsageMode == System.ComponentModel.LicenseUsageMode.Designtime)
{
// Design time logic
}
To find out if you're in "design mode":
Windows Forms components (and controls) have a DesignMode property.
Windows Presentation Foundation controls should use the IsInDesignMode attached property.
The Control.DesignMode property is probably what you're looking for. It tells you if the control's parent is open in the designer.
In most cases it works great, but there are instances where it doesn't work as expected. First, it doesn't work in the controls constructor. Second, DesignMode is false for "grandchild" controls. For example, DesignMode on controls hosted in a UserControl will return false when the UserControl is hosted in a parent.
There is a pretty easy workaround. It goes something like this:
public bool HostedDesignMode
{
get
{
Control parent = Parent;
while (parent!=null)
{
if(parent.DesignMode) return true;
parent = parent.Parent;
}
return DesignMode;
}
}
I haven't tested that code, but it should work.
The most reliable approach is:
public bool isInDesignMode
{
get
{
System.Diagnostics.Process process = System.Diagnostics.Process.GetCurrentProcess();
bool res = process.ProcessName == "devenv";
process.Dispose();
return res;
}
}
The most reliable way to do this is to ignore the DesignMode property and use your own flag that gets set on application startup.
Class:
public static class Foo
{
public static bool IsApplicationRunning { get; set; }
}
Program.cs:
[STAThread]
static void Main()
{
Foo.IsApplicationRunning = true;
// ... code goes here ...
}
Then just check the flag whever you need it.
if(Foo.IsApplicationRunning)
{
// Do runtime stuff
}
else
{
// Do design time stuff
}
I had the same problem in Visual Studio Express 2013. I tried many of the solutions suggested here but the one that worked for me was an answer to a different thread, which I will repeat here in case the link is ever broken:
protected static bool IsInDesigner
{
get { return (Assembly.GetEntryAssembly() == null); }
}
The devenv approach stopped working in VS2012 as the designer now has its own process. Here is the solution I am currently using (the 'devenv' part is left there for legacy, but without VS2010 I am not able to test that though).
private static readonly string[] _designerProcessNames = new[] { "xdesproc", "devenv" };
private static bool? _runningFromVisualStudioDesigner = null;
public static bool RunningFromVisualStudioDesigner
{
get
{
if (!_runningFromVisualStudioDesigner.HasValue)
{
using (System.Diagnostics.Process currentProcess = System.Diagnostics.Process.GetCurrentProcess())
{
_runningFromVisualStudioDesigner = _designerProcessNames.Contains(currentProcess.ProcessName.ToLower().Trim());
}
}
return _runningFromVisualStudioDesigner.Value;
}
}
/// <summary>
/// Are we in design mode?
/// </summary>
/// <returns>True if in design mode</returns>
private bool IsDesignMode() {
// Ugly hack, but it works in every version
return 0 == String.CompareOrdinal(
"devenv.exe", 0,
Application.ExecutablePath, Application.ExecutablePath.Length - 10, 10);
}
System.Diagnostics.Debugger.IsAttached
It's hack-ish, but if you're using VB.NET and when you're running from within Visual Studio My.Application.Deployment.CurrentDeployment will be Nothing, because you haven't deployed it yet. I'm not sure how to check the equivalent value in C#.
using (System.Diagnostics.Process process = System.Diagnostics.Process.GetCurrentProcess())
{
bool inDesigner = process.ProcessName.ToLower().Trim() == "devenv";
return inDesigner;
}
I tried the above code (added a using statement) and this would fail on some occasions for me. Testing in the constructor of a usercontrol placed directly in a form with the designer loading at startup. But would work in other places.
What worked for me, in all locations is:
private bool isDesignMode()
{
bool bProcCheck = false;
using (System.Diagnostics.Process process = System.Diagnostics.Process.GetCurrentProcess())
{
bProcCheck = process.ProcessName.ToLower().Trim() == "devenv";
}
bool bModeCheck = (System.ComponentModel.LicenseManager.UsageMode == System.ComponentModel.LicenseUsageMode.Designtime);
return bProcCheck || DesignMode || bModeCheck;
}
Maybe a bit overkill, but it works, so is good enough for me.
The success in the example noted above is the bModeCheck, so probably the DesignMode is surplus.
You check the DesignMode property of your control:
if (!DesignMode)
{
//Do production runtime stuff
}
Note that this won't work in your constructor because the components haven't been initialized yet.
When running a project, its name is appended with ".vshost".
So, I use this:
public bool IsInDesignMode
{
get
{
Process p = Process.GetCurrentProcess();
bool result = false;
if (p.ProcessName.ToLower().Trim().IndexOf("vshost") != -1)
result = true;
p.Dispose();
return result;
}
}
It works for me.
I'm not sure if running in debug mode counts as real, but an easy way is to include an if statement in your code that checkes for System.Diagnostics.Debugger.IsAttached.
If you created a property that you don't need at all at design time, you can use the DesignerSerializationVisibility attribute and set it to Hidden. For example:
protected virtual DataGridView GetGrid()
{
throw new NotImplementedException("frmBase.GetGrid()");
}
[DesignerSerializationVisibility(DesignerSerializationVisibility.Hidden)]
public int ColumnCount { get { return GetGrid().Columns.Count; } set { /*Some code*/ } }
It stopped my Visual Studio crashing every time I made a change to the form with NotImplementedException() and tried to save. Instead, Visual Studio knows that I don't want to serialize this property, so it can skip it. It only displays some weird string in the properties box of the form, but it seems to be safe to ignore.
Please note that this change does not take effect until you rebuild.
We use the following code in UserControls and it does the work. Using only DesignMode will not work in your app that uses your custom user controls as pointed out by other members.
public bool IsDesignerHosted
{
get { return IsControlDesignerHosted(this); }
}
public bool IsControlDesignerHosted(System.Windows.Forms.Control ctrl)
{
if (ctrl != null)
{
if (ctrl.Site != null)
{
if (ctrl.Site.DesignMode == true)
return true;
else
{
if (IsControlDesignerHosted(ctrl.Parent))
return true;
else
return false;
}
}
else
{
if (IsControlDesignerHosted(ctrl.Parent))
return true;
else
return false;
}
}
else
return false;
}
Basically the logic above boils down to:
public bool IsControlDesignerHosted(System.Windows.Forms.Control ctrl)
{
if (ctrl == null) return false;
if (ctrl.Site != null && ctrl.Site.DesignMode) return true;
return IsControlDesignerHosted(ctrl.Parent);
}
If you are in a form or control you can use the DesignMode property:
if (DesignMode)
{
DesignMode Only stuff
}
I found the DesignMode property to be buggy, at least in previous versions of Visual Studio. Hence, I made my own using the following logic:
Process.GetCurrentProcess().ProcessName.ToLower().Trim() == "devenv";
Kind of a hack, I know, but it works well.
System.ComponentModel.Component.DesignMode == true
To solve the problem, you can also code as below:
private bool IsUnderDevelopment
{
get
{
System.Diagnostics.Process process = System.Diagnostics.Process.GetCurrentProcess();
if (process.ProcessName.EndsWith(".vshost")) return true;
else return false;
}
}
Here's another one:
//Caters only to thing done while only in design mode
if (App.Current.MainWindow == null){ // in design mode }
//Avoids design mode problems
if (App.Current.MainWindow != null) { //applicaiton is running }
After testing most of the answers here, unfortunately nothing worked for me (VS2015).
So I added a little twist to JohnV's answer, which didn't work out of the box, since DesignMode is a protected Property in the Control class.
First I made an extension method which returns the DesignMode's Property value via Reflection:
public static Boolean GetDesignMode(this Control control)
{
BindingFlags bindFlags = BindingFlags.Instance | BindingFlags.NonPublic | BindingFlags.Static;
PropertyInfo prop = control.GetType().GetProperty("DesignMode", bindFlags);
return (Boolean)prop.GetValue(control, null);
}
and then I made a function like JohnV:
public bool HostedDesignMode
{
get
{
Control parent = Parent;
while (parent != null)
{
if (parent.GetDesignMode()) return true;
parent = parent.Parent;
}
return DesignMode;
}
}
This is the only method that worked for me, avoiding all the ProcessName mess, and while reflection should not be used lightly, in this case it did all the difference! ;)
EDIT:
You can also make the second function an extension method like this:
public static Boolean IsInDesignMode(this Control control)
{
Control parent = control.Parent;
while (parent != null)
{
if (parent.GetDesignMode())
{
return true;
}
parent = parent.Parent;
}
return control.GetDesignMode();
}
For WPF (hopefully this is useful for those WPF people stumbling upon this question):
if (System.ComponentModel.DesignerProperties.GetIsInDesignMode(new DependencyObject()))
{
}
GetIsInDesignMode requires a DependencyObject. If you don't have one, just create one.
/// <summary>
/// Whether or not we are being run from the Visual Studio IDE
/// </summary>
public bool InIDE
{
get
{
return Process.GetCurrentProcess().ProcessName.ToLower().Trim().EndsWith("vshost");
}
}
Here's a flexible way that is adaptable to where you compile from as well as whether or not you care which mode you're in.
string testString1 = "\\bin\\";
//string testString = "\\bin\\Debug\\";
//string testString = "\\bin\\Release\\";
if (AppDomain.CurrentDomain.BaseDirectory.Contains(testString))
{
//Your code here
}