I finally found the solution to my problems, and am posting this for anyone who would need some clarification: First, as David pointed out, simply treating the referenced argument as a pointer and assigning it the array’s address does not work. You have to copy the entire array contents into the referenced array. Easily fixed. … Read more
Something like this shall do the trick: $knownIds = array(); foreach( $myArray AS $key=>$item ) { if( array_key_exists($item[‘event_id’], $knownIds) === true ) { unset( $myArray[$key] ); } else { $knownIds[$item[‘event_id’]] = $key; // value does not matter really here } } $myArray = array_values($myArray); where $myArray is your source array. 3 solved Remove From multidimensional … Read more
So you can build out a new javascript object, but it might be tricky if you have object attributes of ‘items’ repeated over in the same object. I fixed this by making the results have an array of items like so. The algorithm is using a double for loop so it would be expensive for … Read more
You must create td and text nodes within loop. This code creates only 2 td, so only 2 are visible. Example: var table = document.createElement(‘table’); for (var i = 1; i < 4; i++) { var tr = document.createElement(‘tr’); var td1 = document.createElement(‘td’); var td2 = document.createElement(‘td’); var text1 = document.createTextNode(‘Text1’); var text2 = document.createTextNode(‘Text2’); … Read more
Use modulus(%) operator to get every 2nd element in loop and add fontStyle on it: var myList = document.getElementById(‘myList’); var addList = [“Python”, “C”, “C++”, “Ruby”, “PHP”, “Javascript”, “Go”, “ASP”, “R”]; for (var i = 0; i < addList.length; i++) { var newLi = document.createElement(“li”); newLi.innerHTML = addList[i]; if(i%2==0){ newLi.style.fontStyle = “italic” newLi.innerHTML = “<strong>”+addList[i]+”</strong>”; … Read more
You need to allocate the vector elements before assigning values to them, when you do vector<int> arr; // defines an empty vector. arr[y] = … // access element at position y. arr here is empty, pointing to a zero array, you are trying to write at memory 0x0000. Checkout vector documentation, to see the different … Read more
Here is a very simple example to get you started. This is for addition (not multiplication or exponentiation), and it’s for 3-digit numbers, not thousands. But it demonstrates the basic idea of holding each digit in one element of an array. When you add (or subtract, or multiply) numbers like these in a computer program, … Read more
You could use the Object.keys() methods to iterate over your object and then use the Array.map() method to transform your object to the array structure you need. var data = { “0”: “value1”, “1”: “value2” } var newData = Object.keys(data).map(key => ({ [key]: data[key] })) console.log(newData) –Update– You would simply need to change the object … Read more
Your reverseArray() method assigns array to the variable revArray, thus the 2 variables pointing the same array in memory. To avoid that, try something like this: public static int[] reverseArray(int[] array){ int[] revArray= Arrays.copyOf(array, array.length); int i=0; int j=revArray.length-1; while(i<=j){ swap(i,j,revArray); j–; i++; } return revArray; } 1 solved Reverse the order of array in … Read more
should try it yourself, but here you go static int [] arr = new int [getcount ()]; public static void main(String[] args) { System.out.println(arr.length); } private static int getcount() { return 7; } output 7 solved Can you have a method call inside of an array declaration?
To understand what is happening under the hood you must understand the pointer and value semantics of for range construct in go. It is clearly explained in this ardan labs article emails := []string{“a”, “b”} CCEmails := []*string{} for _, cc := range emails { p := &cc fmt.Println(cc, p) CCEmails = append(CCEmails,&cc) } The … Read more
There is no reason for giving type mismatch, while you are adding two int array into another int array. It should work: for(int k = 0; k < 20; k++){ EndTime[k] = StarTime[k] + duration[k]; } Ensure that exp.sample() is really casting into int. Also ensure the arrays are of same type. 1 solved Build … Read more
As an example (style and add to it as needed) $cars[] = [‘name’ => ‘HRV’, ‘price’ => ‘CAD-$23300’, ‘img’ => ‘http://direct.automobiles.honda.com/images/2016/hr-v/exterior-gallery-new/2016-honda-hrv-front-view.jpg’, ‘desc’ => ‘HRV is a mini suv made by Honda’, ]; $cars[] = [‘name’ => ‘CHR’ , ‘price’ => ‘CAD-$23675’ , ‘img’ => ‘https://d1bxbrgbmu84na.cloudfront.net/wp-content/uploads/2019/08/16093812/CHR.jpg’ , ‘desc’ => ‘RDX is a large SUV made by … Read more
You can use filter to get the values that occur twice. var arr = [0, 1, 2, 2, 3, 3, 5]; var dups = arr.filter ( (v,i,a) => a.indexOf(v) < i ); console.log(dups); In comments you stated you would only have doubles, but no values that occur more than twice. Note that the above would … Read more
You can try with JSONSimple library and form the object in this way- You have to import org.json.simple.JSONArray and org.json.simple.JSONObject for using this code. JSONObject object=new JSONObject(); JSONObject holder=new JSONObject(); JSONArray taskAssings = new JSONArray(); JSONObject taskAssigned=new JSONObject(); taskAssigned.put(“id”, “3c814009-82f7-4246-bc51-2d263e758561”); JSONObject taskAssignee=new JSONObject(); taskAssignee.put(“id”, “3c814009-82f7-4246-bc51-2d263e758561”); holder.put(“taskAssigned”,taskAssigned); holder.put(“taskAssignee”,taskAssignee); taskAssings.add(holder); object.put(“taskAssings”, taskAssings); JSONObject status=new JSONObject(); status.put(“id”, “7d8a0d80-5c93-46cc-982d-47399503beaa”); … Read more