[ad_1] Yes, the java string is immutable. In changeString, you are passing in a reference to the string lalala and then you are changing the reference to one that points to lalalaCHANGE!!!. The original string object is not changed, and the reference in main still refers to that original object. If you were to use … Read more
[ad_1] echo “<a href=”https://stackoverflow.com/questions/16006445/$href”>$text</a><br />\n”; [ad_2] solved How do I echo the link and it’s text from a scraped page with PHP [closed]
[ad_1] The problem is actually like counting in base sizeof(set) on length k (assuming the set has 10 items maximum). For instance, with a set of { 0, 1, 2 } on length 2, you count from 00 to 22, base 3. To solve the “one digit change only” constraint, instead of counting increasingly, do … Read more
[ad_1] Change your include to #include<string> The string.h contains functions to manipulate string, but not the std::string class. #pragma once Is to prevent headers for being included more then once, leading to duplicated symbols. In C++, #include means the compiler just replaces the #include with the contents of the file included. Imagine you have A.h … Read more
[ad_1] For this you don’t even need regex. Just split the String and call the right indices: String latLongString = “Tracking info Latitude: 3.9574667 Longitude: 7.44882167″; //yours obviously won’t be hardcoded String[] split = latLongString.split(” “); String lat = split[3]; String lon = split[5]; [ad_2] solved Extracting two numbers from a string using java
[ad_1] Is this what you are looking for? string[,] prop; //This is a 2D string List<List<string>> mysteryList; if (value is object[,]) { object[,] objArray = (object[,])value; // Get upper bounds for the array int bound0 = objArray.GetUpperBound(0);//index of last element for the given dimension int bound1 = objArray.GetUpperBound(1); prop = new string[bound0 + 1, bound1 … Read more
[ad_1] You’re outputting “hundred” instead of “one hundred”. 6 [ad_2] solved C++ character count (Project Euler 17 wrong answer) [closed]
[ad_1] This one should be the way: #(\w+)(?:[, .]|$) # matches # literally \w is a word with at least one letter (?:) is non-capturing group [, .]|$ is set of ending characters including the end of line $ For more information check out Regex101. In Java don’t forget to escape with double \\: String … Read more
[ad_1] Set<String> words = new HashSet<>(); String str = “Java Ruby PHP. Java is good. PHP please looks at Java”; Matcher mat = Pattern.compile(“\\b(\\w+)(?=\\b.*\\1)”).matcher(str); while(mat.find()){ words.add(mat.group(1)); } System.out.println(words); Also you can just split string by words and calculate count using Map. And then filter words with count > 1. 1 [ad_2] solved Trouble with finding … Read more
[ad_1] “I do not need a solution with saying me to make my own class. I definitely will not re-invent an already existing data-type!” List<string,string,string> is not an already existing data type. You probably did something like this in the past; // Another simple way would be to create a class which has a constructor … Read more
[ad_1] Returning code from server is terrible awful. Anyways, this code I give you do some steps: removes new lines and multispaces into one length space from input. finds words by pattern where word starts with if and ends with digits. Adds them to array. prepares pattern with user chosen options combining into similar to … Read more
[ad_1] Okay so, I don’t really think there is any error, but I will propose you a modified version of your code, which will be more readable, more efficient, and less error prone: // #include<bits/stdc++.h> I don’t know what that include is, use more specific header based on your needs like #include <string> // for … Read more
[ad_1] Let’s start with this string : let abc : String = “aaaabbbbbbbccddde” And have the output in a new variable var result = “” Let’s use an index to go through the characters in the string var index = abc.startIndex while index < abc.endIndex { //There is already one character : let char = … Read more
[ad_1] string str = “12345678901234”; string str1 = string.Empty; string str2 = string.Empty; string str3 = string.Empty; for (int i = 0; i < str.Length; i++) { if (i < 2) str1 += str[i]; else if (i > 1 && i < 8) str2 += str[i]; else str3 += str[i]; } 5 [ad_2] solved Extract … Read more
[ad_1] You are confusing strings and characters. A character literal is enclosed in ‘. A string literal is enclosed in ” and is an array of characters. e.g.: “This is a string literal” ‘c’ // this is character To test if a char is not z for instance, you write: char c = …; if … Read more