The function does not make sense. For starters it is unclear why you are using a pointer to the vector instead of a reference to the vector. Nevertheles, this declaration vector<unsigned char> vec(5); does not reseeve a memory for the vector. It initializes the vector with 5 zero-characters which will be appended with other characters … Read more
if you need to count characters in string, try the following a = “aaabbcccd” b = dict.fromkeys(a, 0) for i in a: b[i] += 1 b now holds the counts you require: {‘a’: 3, ‘c’: 3, ‘b’: 2, ‘d’: 1} solved Counting the occurrence of unique letters in a string in Python 3
Have you tried calculating the length within the function rather than in the console log? This ensures that all of the logic is first completed within the function before it is returned in the console log. Your function expects the string name, but you are passing name.length function getNameLength(name){ return name.length; } console.log(getNameLength(‘John’)); console.log(getNameLength(‘Argentina!’)); console.log(getNameLength(‘Macedonia’)); … Read more
char a[50][50] declares a as a 50-element array of 50-element arrays of char. That means each a[i] is a 50-element array of char. It will be laid out in memory like: +—+ a: | | a[0][0] +—+ | | a[0][1] +—+ | | a[0][2] +—+ … +—+ | | a[0][49] +—+ | | a[1][0] +—+ … Read more
Here is a description of malloc(). It takes a parameter size, in bytes. So if the user entered 5 words, then you need to first allocate an array big enough to store 5 pointers. Eg. char ** words = NULL; words = malloc(sizeof(char *) * <NUMBER OF CHARACTERS THE USER ENTERED>); If we assume ASCII … Read more
try { String formatted = String.valueOf((int)Double.parseDouble(“12345.0”)); } catch (ParseException e) { // input is not a number } 0 solved How to format this string in java?
Introduction This tutorial will provide a step-by-step guide on how to use jQuery to find a word in a string and wrap it with a tag. This is a useful technique for highlighting certain words or phrases in a string of text. We will be using the jQuery .each() and .wrap() functions to achieve this. … Read more
You need to do this : var txt = textDiv.replace(new RegExp(searchInput, ‘gi’), function(str) { return “<span class=”highlight”>” + str + “</span>” }); or var txt = textDiv.replace( new RegExp(searchInput, ‘gi’), “<span class=”highlight”>$&</span>”); gi -> all case insensitive matching text $(document).ready(function() { // globals var searchInput; var textDiv; $(‘.searchBtn’).click(function(event) { event.preventDefault(); // set the values of … Read more
Can anyone explain why address operator is returning string and not the address in the following C++ program? [closed] solved Can anyone explain why address operator is returning string and not the address in the following C++ program? [closed]
You’re reusing the same string buffer. If you keep putting things into the same buffer without clearing it, you’re obviously going to get extraneous stuff from previous iterations. Simply declare the StringBuffer inside the while loop so that it is created on each iteration. Anyway, you should learn to use your debugger, instead of asking … Read more
Yes it’s possible: String test = “voiceemailchat”; String find = “chat”; if(test.contains(find)) { //string found } 0 solved Search a word in java?
Something like this… List<String> result = Stream.of(“A”, “A”, “BB”, “C”, “BB”, “D”) .collect(Collectors.groupingBy( Function.identity(), Collectors.counting())) .entrySet() .stream() .filter(x -> x.getValue() == 1L) .map(Entry::getKey).collect(Collectors.toList()); System.out.println(result); // [C, D] 2 solved return the list with strings with single occurrence [closed]
First make ‘path’ a StringBuffer to make it easier to work with. Then use indexes: StringBuffer path = new StringBuffer(“cmd /c start D:\\SMPPPushGW_SMSCID1_Passive\\note.bat”); String result = path.substring(path.indexOf(“start”)+5, path.lastIndexOf(“\\”)); As mentioned, File API may be useful, but also URI. solved Working with strings in java : extract a particular string [closed]
You need simply enforce result string to show more zeroes if need. var a = 20.0f; a.ToString(“00.0000”, CultureInfo.InvariantCulture) //2 digits before and 4 digits after (.) Pay attention on fact that in your case the value may not be exactly 20.0, but something like 20.0012. In these cases you need to first convert it to … Read more
String.trim() method remove trailing and leading white spaces from the string. but since the string you are testing doesn’t have any trailing or leading white spaces, the below two will certainly return the same string. str.replaceFirst(“(.*)<?xml”,”<?xml”); str.trim().replaceFirst(“(.*)<?xml”,”<?xml”) However, your regular expression only removes leading white spaces, so if the testing string has trailing white spaces … Read more