They check completely different things. is compares types. == compares values. var isString = “Abc” is String; // => true var equalToString = “Abc” == String; // Error: `string’ is a `type’ but a `variable’ was expected There is one domain where these can both apply, and have different meanings, and that’s in type checking: … Read more
You need to define that constant in MainActivity as public static final String EXTRA_MESSAGE = “extra_message”; so that in DisplayMessageActivity you can access that static constant as String message = intent.getStringExtra(MainActivity.EXTRA_MESSAGE); MainActivity public class MainActivity extends AppCompatActivity { public static final String EXTRA_MESSAGE = “extra_message”; // static : accessed by class name // final : … Read more
How can I remove every digit at the beginning? Linq SkipWhile would be one way string result = string.Concat(“1234ABC123”.SkipWhile(char.IsDigit)); // “ABC123” solved Check if the first chars are digits
We have: 1 : if statement 3 : * operator 3 : function statement Totally: 7 So we have: T(n)=t(n-1) + t(n-2) + t(n-3) + 7 T(99)=1 T(98)=1 … T(1)=1 Then to reduce T(n), we have T(n-3)<T(n-2)<T(n-1) therefore: T(n) <= T(n-1)+T(n-1)+T(n-1) + 7 T(n) <= 3T(n-1) + 7 So we can solve T(n) = 3T(n-1) … Read more
First, let’s use a typedef for the function: typedef int agechanger(int); this makes a new type, agechanger, which will be used in code for passing the function instances around. Now, you should give your person class a proper constructor, and properly incapsulate the age field providing a public getter. Then add a method that accepts … Read more
Use output.put(i); since it takes one character at a time. But you really want to skip the for-loop and just write the whole vector in one write: output.write(f.data(), f.size()); 2 solved Save an unsigned char vector one by one
1. Thanks a lot for all your support guys 2. Finally I found the solution. ———- package demo.practice.java; import java.util.Scanner; public class Demo { public static void main(String[] args) { System.out.println(“Enter String”); Scanner sc=new Scanner(System.in); String input=sc.nextLine(); //split into words String[] words = input.split(“(?=[A-Z])”); words[0] = capitalizeFirstLetter(words[0]); //join StringBuilder builder = new StringBuilder(); for ( … Read more
If you want to add a a value to key in your dictionary just make sure the values are, from the start, tuples or lists. Once you do this you can just: d = {1:(0,),2:(0,)} d[1] = d[1]+(0,) And a new value is now present at d[1]. Other than that and you might have to … Read more
This will always return 0, because the last fac will always be 0 and it multiplies all the other results. Change that to if(a<1) fac=1; Also indent your code properly. solved what is wrong with my factorial program ??? [closed]
The code you need will be something like this: <head> <script> function play(){ var myAudio = document.getElementById(“audio”); if(myAudio.paused) myAudio.play(); else myAudio.pause(); } </script> </head> <body> <img src=”https://stackoverflow.com/questions/35963057/SoundWave.gif” width=”200″ height=”200″ onclick=”play();”> <audio id=”audio” src=”01.mp3″></audio> </body> you can get more information about html5 audio element and how to work with it using this link P.S. <img> tag … Read more
public static Object nthToLast(int n, LinkedList list) { int len = list.size(); if(n<0 || len < n){ return null; } return list.get(len-n); } 1 solved Check my solution for How to find nth element from the end of a singly linked list [closed]
I think you need something like this: // Make sure x is smaller than y if(x>y) { swap(&x, &y); } // Make sure x is smaller than z if(x>z) { swap(&x, &z); } // Now x is smaller than both y and z // Make sure y is smaller than z if(y>z) { swap(&y, &z); … Read more
My guess is that it’s a numpy array, not a list. This is logical indexing. The > comparison returns an array of booleans. Wherever those are True, the corresponding element of img is set to [0, 0, 255]. More directly, it’s creating a ring of blue points, where the radius of the inner empty circle … Read more
The simple way to get a count of unique words is to use a set. I put your text into a file called ‘qdata.txt’. The file is very small, so there’s no need to read it line by line: just read the whole thing into a single string, then split that string on whitespace and … Read more
You defined one and only one object when you executed: tino=read.delim(“clipboard”) The column names of that object are not handled as other objects. If you wnated to create a new object from that dataframe you could do this: x <- with(tino, cbind(DEX, GRW , Debt, Life) ) It’s possible this might do violence to the … Read more