![]() ![]() Endindex of the substring starts from 1 and not from 0. Replaces each substring of this string that matches the literal target sequence with the specified literal replacement sequence. The substring begins with the character at the specified index and extends to the end of this string. How long would you say are your strings when you get a 10x speed improvement? Just so I have an idea what lenghts we’re talking about. String substring () The substring () method has two variants and returns a new string that is a substring of this string. I just finished my final exam for this semester and I wanted to take a look at JMH for a while now. To solve this, also use Pattern.quote as suggested in the comments. Similar to StringBuilder. 10 Answers Sorted by: 356 String target 'FOOBar' target target.replaceAll (' (i)foo', '') (target) Output: Bar It's worth mentioning that replaceAll treats the first argument as a regex pattern, which can cause unexpected results. I might be able to write a benchmark tomorrow. Using StringBuffer.replace() StringBuffer is a mutable sequence of characters. 6 and 8 are the 2 bytecode targets for kotlin so this is interesting to compare and important so that we don’t have performance problems (especially since many android devices still use java 6, I think) and the newer jvm versions for the improved string implementations. The first parameter is the substring to be. Also I think it would be helpful to have actual benchmark tests, with the current implementation, a new kotlin implementation (like the one above) and the java implementation for different string lengths on jvm version 6, 8, 9, and 13. The first and most commonly used method to remove/replace any substring is the replace() method of Java String class. If you want to access the last two indices of a string, then you can use: - str.substring(str.length() - 2) This gives you string from index str.length() - 2 to the last character, which is exactly the last two character. I think the next step to solve this would be to create an issue at. To replace the nth Occurrence String we have developed 2 methods in the above example:- replaceNthOccurance() and indexOfNthOccurrence(). That said, the current implementation is slow for long strings as you pointed out. Before java 9 the java implementation used regex which comes with it’s own performance problems, so this could be the reason that kotlin didn’t use the jvm implementation. That said depending on what java version is used might be faster or slower (depending on the input text). I’m aware that your problem is with the kotlin implementation. An invocation of this method of the form str. localStringBuilder.toString ().replace (' sender', callerName) will work exaclty as you expect, because it takes normal Strings as both parameters. ![]() are important parts of regexes using to group expressions. ![]() My question was in regards to test, since he couldn’t verify your results. Replaces each substring of this string that matches the given regular expression with the given replacement. replaceAll take REGEX as input, not a String, but regex. ![]()
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