Sorry to completely rewrite this post, but I was way off on my troubleshooting. Hopefully this prevents someone's headache in the future...
Now, for example code.
function compare(uint8 a, uint8 b)
private
requireUnpaused
returns(bool)
{
// increment a by 1
a = a++;
// if a + 1 = 3, then loop it around the cycle to be 0
if (a >= 3) {
a = 0;
}
// compare a to b. If a = b, a is the winner - return true
return a == b ? true : false;
}
This code works as expected if I replace "a = a++" with "a = ++a", just "a++" or "++a", or even "a = a + 1".
This drove me crazy for a few days. My question for someone more versed in this than I: Why does "a = a++" not work, when every other way of incrementing "a" by 1 does?
Related
I'm kind of stuck, I don't know how to make the second loop to start 1 position above the first loop in Kotlin.
I have an array (named myArray) with 10 elements, I need to Write a Kotlin program that prints the number that has the most consecutive repeated number in the array and also prints the number of times it appears in the sequence.
The program must parse the array from left to right so that if two numbers meet the condition, the one that appears first from left to right will be printed.
Longest: 3
Number: 8
fun main() {
val myArray: IntArray = intArrayOf(1,2,2,4,5,6,7,8,8,8)
for((index , value) in myArray.withIndex()){
var inx = index + 1
var count = 0
var longest = 0
var number = 0
for((inx,element) in myArray.withIndex()) {
if(value == element ){
count+=
}
}
if(longest < count){
longest = count
number = value
}
}
}
I'm against just dropping answers, but it is quite late for me, so I'll leave this answer here and edit it tomorrow with more info on how each part works. I hope that maybe in the meanwhile it will help you to gain some idea to where you might be going wrong.
val results = mutableMapOf<Int, Int>()
(0..myArray.size - 2).forEach { index ->
val current = myArray[index]
if (current == myArray[index + 1]) {
results[current] = (results[current] ?: 1) + 1
}
}
val (max, occurrences) = results.maxByOrNull { it.value } ?: run { println("No multiple occurrences"); return }
println("Most common consecutive number $max, with $occurrences occurrences")
Alternatively if the intArray would be a list, or if we allowed to change it to a list myArray.toList(), you could replace the whole forEach loop with a zipWithNext. But I'm pretty sure that this is a HW question, so I doubt this is the expected way of solving it.
myList.zipWithNext { a, b ->
if (a == b) results[a] = (results[a] ?: 1) + 1
}
Hey guys I'm new to Game Maker Studio and new to the language. I'm making a game and have been working on the dialogue system.
This chunk of code was designed for characters respond to a set of choices, the dialogue starts by printing out the first element of the line_array, which it does, then give the player the choice of two responses from the response_array, which it insteads prints out the second element of the line_array and I don't understand why.
Does an argument only hold one element of an array? I'm initializing two arrays in an object oCivilian2 and pushing them through code DialogueCode which is linked to another object oRespond that supposed to allow me to sift through dialogue in game. Anything helps thanks
It's initialized here in create of oCivilian2
line_array = [3];
line_array[0] = "Ethan it's good to see you! \n I thought after the incident well.... \n well I thought we had lost you";
line_array[1] = "I've said too much";
line_array[2] = "You hit your head trying to saver her\n It was horrible";
response_array = [2];
response_array[0] = "What happened?";
response_array[1] = "I don't recall alot. How bad was it?";
counter = 0;
x1 = RESOLUTION_W / 2;
y1 = RESOLUTION_H -70;
x2 = RESOLUTION_W/2;
y2 = RESOLUTION_H;
_print = "";
responseSelected = 0;
Then the step which links it to DialogueCode when spacebar is pressed
keyActivate = keyboard_check_pressed(vk_space);
if (keyActivate)
{
var inst = collision_rectangle(oPlayer.x+3,oPlayer.y+3,oPlayer.x-3,oPlayer.y-3, oCivilian2, false, false);
if (inst != noone)
{
ScriptExecuteArray(DialogueCode, line_array);
ScriptExecuteArray(DialogueCode, response_array);
}
}
Then through to step in the object oRespond
lerpProgress += (1 - lerpProgress) / 50;
textProgress += global.textSpeed;
x1 = lerp(x1, x1Target,lerpProgress);
x2 = lerp(x2, x2Target,lerpProgress);
keyUp = (keyboard_check_pressed(vk_up)) || (keyboard_check_pressed(ord("W")))
keyDown = keyboard_check_pressed(vk_down) || keyboard_check_pressed(ord("S"));
responseSelected += (keyDown - keyUp);
var _max = 2;
var _min = 0;
if (responseSelected > _max) responseSelected = _min;
if (responseSelected < _min) responseSelected = _max;
for (var i = 0; i < 2; i++)
{
var _marker = string_pos(",", response);
if (string_pos(",",response))
{
responseScript[i] = string_copy(response,0,_marker);
string_delete(response,0,_marker);
var _marker = string_pos(",", response);
}
else
{
responseScript[i] = string_copy(response,0, string_length(response));
}
}
if (keyboard_check_pressed(vk_space))
{
counter++;
}
Then to print in oRespond
/// text
//response
NineSliceBoxStretched(sTextBox, x1,y1,x2,y2, 0);
draw_set_font(fText);
draw_set_halign(fa_center);
draw_set_valign(fa_top);
draw_set_color(c_black);
if (counter % 2 == 0)
{
var _i = 0;
var _print = string_copy(text,1,textProgress);
draw_text((x1+x2) / 2, y1 + 8, _print);
draw_set_color(c_white);
draw_text((x1+x2) / 2, y1 + 7, _print);
_i++;
}
else
{
if (array_length_1d(responseScript) > 0)
{
var _print = "";
for (var t = 0; t < array_length_1d(responseScript); t++)
{
_print += "\n";
if (t == responseSelected) _print += "--> "
_print += responseScript[t];
show_debug_message(responseScript[t]);
if (t == responseSelected) _print += " <-- "
}
draw_text((x1+x2) / 2, y1 + 8, _print);
draw_set_color(c_white);
draw_text((x1+x2) / 2, y1 + 7, _print);
}
}
Alright, i think to see many problems with your code.
First of all, since arrays in GM are dynamic declare them like
line_array[3]
is a bad practice (in my point of view)
I've never declared an array this way in GM so that could be the problem here.
Second, i don't really understand the logic of your code, always create objects, at least in the GM environment, that corresponds to "physical" entities, i would make an object for the Civilian but not for the "respond".
I've red your code a lot of times and since no one answered you in 3 months i can assume it's because no one can really understand your way of coding, and this way of coding will probably give you a lot of problems in future. The thing that you're trying to doing could be super-easy if done with a good hierarchy.
I would like to help u with this code, but i find it very chaotic.
If you've not resolved this problems, write a comment :)
I advice you to fully re-implement it even if resolved anyway.
Here's the problem in which I encountered this issue:
The function should compare the value at each index position and score a point if the value for that position is higher. No point if they are the same. Given a = [1, 1, 1] b = [1, 0, 0] output should be [2, 0]
fun compareArrays(a: Array<Int>, b: Array<Int>): Array<Int> {
var aRetVal:Int = 0
var bRetVal:Int = 0
for(i in 0..2){
when {
a[i] > b[i] -> aRetVal + 1 // This does not add 1 to the variable
b[i] > a[i] -> bRetVal++ // This does...
}
}
return arrayOf(aRetVal, bRetVal)
}
The IDE even says that aRetVal is unmodified and should be declared as a val
What others said is true, but in Kotlin there's more. ++ is just syntactic sugar and under the hood it will call inc() on that variable. The same applies to --, which causes dec() to be invoked (see documentation). In other words a++ is equivalent to a.inc() (for Int or other primitive types that gets optimised by the compiler and increment happens without any method call) followed by a reassignment of a to the incremented value.
As a bonus, consider the following code:
fun main() {
var i = 0
val x = when {
i < 5 -> i++
else -> -1
}
println(x) // prints 0
println(i) // prints 1
val y = when {
i < 5 -> ++i
else -> -1
}
println(y) // prints 2
println(i) // prints 2
}
The explanation for that comes from the documentation I linked above:
The compiler performs the following steps for resolution of an operator in the postfix form, e.g. a++:
Store the initial value of a to a temporary storage a0;
Assign the result of a.inc() to a;
Return a0 as a result of the expression.
...
For the prefix forms ++a and --a resolution works the same way, and the effect is:
Assign the result of a.inc() to a;
Return the new value of a as a result of the expression.
Because
variable++ is shortcut for variable = variable + 1 (i.e. with assignment)
and
variable + 1 is "shortcut" for variable + 1 (i.e. without assignment, and actually not a shortcut at all).
That is because what notation a++ does is actually a=a+1, not just a+1. As you can see, a+1 will return a value that is bigger by one than a, but not overwrite a itself.
Hope this helps. Cheers!
The equivalent to a++ is a = a + 1, you have to do a reassignment which the inc operator does as well.
This is not related to Kotlin but a thing you'll find in pretty much any other language
So I'm trying to figure out if that blue line is in the right place, I know that I should have 9 edges but not sure if it's correct.
The code
public int getResult(int p1, int p2) {
int result = 0; // 1
if (p1 == 0) { // 2
result += 1; //3
} else {
result += 2; //4
}
if (p2 == 0) { //5
result += 3; //6
} else {
result += 4; //7
}
return result; //8 exit node
}
so 8 nodes and it should have 9 edges, right? Did I do the right thing?
Yes, the blue line is placed correctly because after the 3rd line, your program is going to jump to the 5th line.
The easiest way to compute cyclomatic complexity without drawing any flow diagram is as follows:
Count all the loops in the program for, while, do-while, if. Assign a value of 1 to each loop. Else should not be counted here.
Assign a value of 1 to each switch case. Default case should not be counted here.
Cyclomatic complexity = Total number of loops + 1
In your program, there are 2 if loops, so the cyclomatic complexity would be 3(2+1)
You can cross-check it with the standard formulae available as well which are as below:
C = E-N+2 (9-8+2=3)
OR
C = Number of closed regions + 1 (2+1=3)
According to wikipedia:
M = E ā N + 2P,
where
E = the number of edges of the graph.
N = the number of nodes of the graph.
P = the number of connected components.
so:
9 - 8 + 2*1 = 3
So i've tried interpreting this pseudocode a friend made and i wasn't exactly sure that my method returns the right result. Anyone who's able to help me out?
I've done some test cases where e.g. an array of [2,0,7] or [0,1,4] or [0, 8, 0] would return true, but not cases like: [1,7,7] or [2,6,0].
Array(list, d)
for j = 0 to dā1 do
for i = 0 to dā1 do
for k = 0 to dā1 do
if list[j] + list[ i] + list[k] = 0 then
return true
end if
end for
end for
end for
return false
And i've made this in java:
public class One{
public static boolean method1(ArrayList<String> A, int a){
for(int i = 0; i < a-1; i++){
for(int j = 0; j < a-1; j++){
for(int k = 0; k < a-1; k++){
if(Integer.parseInt(A.get(i)+A.get(j)+A.get(k)) == 0){
return true;
}
}
}
}
return false;
}
}
Thanks in advance
For a fix to your concrete problem, see my comment. A nicer way to write that code would be to actually use a list of Integer instead of String, because you will then want to convert the strings back to integers. So, your method looks better like this:
public static boolean method(List<Integer> A) {
for (Integer i : A)
for (Integer j : A)
for (Integer k : A)
if (i + j + k == 0)
return true;
return false;
}
See that you don't even need the size as parameter, since any List in Java embeds its own size.
Somehow offtopic
You're probably trying to solve the following problem: "Find if a list of integers contains 3 different ones that sum up to 0". The solution to this problem doesn't have to be O(n^3), like yours, it can be solved in O(n^2). See this post.
Ok, so here is what I believe the pseudo code is trying to do. It returns true if there is a zero in your list or if there are three numbers that add up to zero in your list. So it should return true for following test cases. (0,1,2,3,4,5), (1,2,3,4,-3). It will return false for (1,2,3,4,5). I just used d=5 as a random example. Your code is good for the most part - you just need to add the ith, jth and kth elements in the list to check if their sum equals zero for the true condition.