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How to “perfectly” override a dict?
...or other builtins) directly. It often makes no sense, because what you actually want to do is implement the interface of a dict. And that is exactly what ABCs are for.
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How to Remove ReadOnly Attribute on File Using PowerShell?
...e here? As for Keith's answer, he's using PSCX. Not everyone has those installed and that's not really a case of PowerShell v1 vs. v2.
– Joey
Apr 22 '11 at 9:28
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Passing references to pointers in C++
As far as I can tell, there's no reason I shouldn't be allowed to pass a reference to a pointer in C++. However, my attempts to do so are failing, and I have no idea why.
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Case-insensitive string comparison in C++ [closed]
...case-insensitive string comparison in C++ without transforming a string to all uppercase or all lowercase?
30 Answers
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Invert “if” statement to reduce nesting
...cked this code from the refactoring catalog. This specific refactoring is called: Replace Nested Conditional with Guard Clauses.
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Python Pandas merge only certain columns
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You can use .loc to select the specific columns with all rows and then pull that. An example is below:
pandas.merge(dataframe1, dataframe2.iloc[:, [0:5]], how='left', on='key')
In this example, you are merging dataframe1 and dataframe2. You have chosen to do an outer left jo...
How to call an async method from a getter or setter?
...ll get populated without blocking the UI, when getTitle() returns.
string _Title;
public string Title
{
get
{
if (_Title == null)
{
Deployment.Current.Dispatcher.InvokeAsync(async () => { Title = await getTitle(); });
}
return _Title;
}
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How can I fix the Microsoft Visual Studio error: “package did not load correctly”?
I installed Visual Studio 2012 and DevExpress 13.1. As Visual Studio started, it generated an error shown by this attached image,
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Android, How can I Convert String to Date?
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answered Nov 30 '17 at 9:32
paolixxpaolixx
1133 bronze badges
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What would cause an algorithm to have O(log n) complexity?
...er dividing n in half i times we'll have a value that is at most 1. The smallest i for which this is guaranteed is roughly log2 n, so if we have an algorithm that divides by 2 until the number gets sufficiently small, then we can say that it terminates in O(log n) steps.
An important detail is tha...