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std::wstring VS std::string
I am not able to understand the differences between std::string and std::wstring . I know wstring supports wide characters such as Unicode characters. I have got the following questions:
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How do you use String.substringWithRange? (or, how do Ranges work in Swift?)
I have not yet been able to figure out how to get a substring of a String in Swift:
33 Answers
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SQLiteDatabase.query method
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tableColumns
null for all columns as in SELECT * FROM ...
new String[] { "column1", "column2", ... } for specific columns as in SELECT column1, column2 FROM ... - you can also put complex expressions here:
new String[] { "(SELECT max(column1) FROM table1) AS max" } would give you a colu...
How to insert newline in string literal?
In .NET I can provide both \r or \n string literals, but there is a way to insert
something like "new line" special character like Environment.NewLine static property?
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How to resolve “must be an instance of string, string given” prior to PHP 7?
...s. Scalar types are not type-hintable. In this case an object of the class string is expected, but you're giving it a (scalar) string. The error message may be funny, but it's not supposed to work to begin with. Given the dynamic typing system, this actually makes some sort of perverted sense.
You ...
Passing a String by Reference in Java?
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You have three options:
Use a StringBuilder:
StringBuilder zText = new StringBuilder ();
void fillString(StringBuilder zText) { zText.append ("foo"); }
Create a container class and pass an instance of the container to your method:
public class Containe...
How to change an Android app's name?
...reen Class name.
Please make sure that you change label:
android:label="@string/title_activity_splash_screen"
in your Splash Screen activity in your strings.xml file. It can be found in Res -> Values -> strings.xml
See more here.
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Declare and initialize a Dictionary in Typescript
... you would expect: Index signatures are
incompatible. Type '{ firstName: string; }' is not assignable to type
'IPerson'. Property 'lastName' is missing in type '{ firstName:
string; }'.
Apparently this doesn't work when passing the initial data at declaration.
I guess this is a bug in TypeS...
Android: How do I get string from resources using its name?
I would like to have 2 languages for the UI and separate string values for them in my resource file res\values\strings.xml :
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Android Split string
I have a string called CurrentString and is in the form of something like this
"Fruit: they taste good" . I would like to split up the CurrentString using the : as the delimiter. So that way the word "Fruit" will be split into its own string and "they taste good" will be another strin...