C does not and never has had a native string type. By convention, the language uses arrays of char terminated with a null char, i.e., with ‘\0′. Functions and macros in the language’s standard libraries provide support for the null-terminated character arrays, e.g., strlen iterates over an array of char until it encounters a ‘\0’ …
How about this? print ”’ dog cat ”'[1:-1] Or so long as there’s no indentation on the first line or trailing space on the last: print ”’ dog cat ”’.strip() Or even, if you don’t mind a bit more clutter before and after your string in exchange for being able to nicely indent it: from …
The pipe, |, is a special-character in regex (meaning “or”) and you’ll have to escape it with a \. Using your current regex: \|.*$ I’ve tried this in Notepad++, as you’ve mentioned, and it appears to work well.
There is a way to ““deaccent” the string being compared” without the use of a substitution function that lists all the accents you want to remove… Here is the easiest solution I can think about to remove accents (and other diacritics) from a string. See it in action: var string = ‘Ça été Mičić. ÀÉÏÓÛ’; …
An easy one as an extension method public static class Extensions { public static Stream ConvertToBase64(this Stream stream) { byte[] bytes; using (var memoryStream = new MemoryStream()) { stream.CopyTo(memoryStream); bytes = memoryStream.ToArray(); } string base64 = Convert.ToBase64String(bytes); return new MemoryStream(Encoding.UTF8.GetBytes(base64)); } }
You can use replaceFirst(String regex, String replacement) method of String.
write % twice: rate = 99.99 let str = String(format: “%.2f%%”, rate)
Use re.search(): >>> import re >>> s1 = “thishasadigit4here” >>> m = re.search(r”\d”, s1) >>> if m: … print(“Digit found at position”, m.start()) … else: … print(“No digit in that string”) … Digit found at position 13
var firstLine = theString.split(‘\n’)[0];
If you have just a pair of brackets ( [] ) in your string, you can use indexOf(): String str = “ABC[ This is the text to be extracted ]”; String result = str.substring(str.indexOf(“[“) + 1, str.indexOf(“]”));