i am still not good coder, please give me advice how to init, read, write Variable-name store in Array MQL4.
the objective, i want to try make EA multi-pair in single windows and keep ALL-STATUS in entire loop process like code below:.
How to simplify the :
if (pairs[cmp] == "EURUSD")
{
VAR1 = str01EURUSD;
VAR2 = int01EURUSD;
VAR3 = date01EURUSD;
VAR4 = dou01EURUSD;
.....
.....
VARN = xxx01EURUSD;
}
if (symbolx == "EURUSD")
{
str01EURUSD = VAR1;
int01EURUSD = VAR2;
date01EURUSD = VAR3;
dou01EURUSD = VAR4;
.....
.....
xxx01EURUSD = VARN
}
in Array Manipulation. Because if there is additional status-code, it will takes time and increasing the row.
SAMPLE CODE :
//multipair
string pairs[] =
{
"EURUSD",
"GBPUSD",
"AUDUSD",
"USDCAD",
"CHFJPY",
.........
.........
"XAUUSD"
};
//Variable name pair
string str01EURUSD = "BUY";
..
..
string str01XAUUSD = "BUY";
int int01EURUSD = 0;
..
..
int int01XAUUSD = 0;
datetime date01EURUSD;
..
..
datetime date01XAUUSD;
double dou01EURUSD = 0;
..
..
double dou01XAUUSD = 0;
int start()
for(int cmp = 0; cmp < ArraySize(pairs); cmp++)
{
if (pairs[cmp] == "EURUSD")
{
VAR1 = str01EURUSD;
VAR2 = int01EURUSD;
VAR3 = date01EURUSD;
VAR4 = dou01EURUSD;
.....
.....
VARN = xxx01EURUSD;
}
.............
.............
if (pairs[cmp] == "XAUUSD")
{
VAR1 = str01XAUUSD;
VAR2 = int01XAUUSD;
VAR3 = date01XAUUSD;
VAR4 = dou01XAUUSD;
.....
.....
VARN = xxx01XAUUSD;
}
// call function checkdata
checkdata(pair[cmp], VAR1, VAR2, VAR3, VAR4, VARN);
if (symbolx == "EURUSD")
{
str01EURUSD = VAR1;
int01EURUSD = VAR2;
date01EURUSD = VAR3;
dou01EURUSD = VAR4;
.....
.....
xxx01EURUSD = VARN
}
if (symbolx == "XAUUSD")
{
str01XAUUSD = VAR1;
int01XAUUSD = VAR2;
date01XAUUSD = VAR3;
dou01XAUUSD = VAR4;
.....
.....
xxx01XAUUSD = VARN
}
}
//checkdata function
checkdata(string symbol, string code01, int code02, datetime code03, double code04, ....., ....., xxxx codeN)
{
all checking process ...
............
............
symbolx = symbol;
VAR1 = code01;
VAR2 = code02;
VAR3 = code03;
VAR4 = code04;
....
....
VARN = codeN;
return(symbolx);
return(VAR1);
return(VAR2);
return(VAR3);
return(VAR4);
.....
.....
return(VARN);
}
Use classes that inherit CObject.mqh and put them all into CArrayObj.mqh list. Do initialization (sth like pairs[cmp] == "EURUSD") { VAR1 = str01EURUSD; ... } by filling the fields of the class instance then add classes to list in OnInit() then use the list to loop over params in OnTick()
Sample code below
#include <Arrays/ArrayObj.mqh>
class CParams : public CObject
{
string m_symbol;
datetime m_date01;
double m_double01;
bool checkData()
{
//do not know what you may need here, return true if need to proceed;
// or signature change to int that returns -1 (SELL), 0 (pass) or 1(BUY)
return false;
}
public:
CParams(const string smb,const datetime dt01,const double d01):m_symbol(smb),m_date01(dt01),m_double01(d01){}
~CParams(){}
void onTick()
{
if(!checkData())
return;
//open another trade?
}
};
CArrayObj *params;
int OnInit()
{
params=new CArrayObj();
params.Add(new CParams("EURUSD",date01EURUSD,dou01EURUSD));
// add more instances of CParams
return(INIT_SUCCEEDED);
}
void OnDeinit(const int reason)
{
delete(params);
}
void OnTick()
{
for(int i=params.Total()-1;i>=0;i--)
{
CParams *param=params.At(i);
param.onTick();
}
}
Related
I'm new to Dart. I was trying to convert integer to roman. But It returns nothing. Can you guys help me? here is my code sample.
this code is from the Leetcode problem section.
class Solution {
String intToRoman(int num) {
List<int> numbers = [1,4,5,9,10,40,50,90,100,400,500,900,1000];
List<String> romans = ["I","IV","V","IX","X","XL","L","XC","C","CD","D","CM", "M"];
int index = romans.length - 1;
String roman = '';
for(num >0;numbers[index]<=num;){
roman += romans[index];
num -= numbers[index];
index -= 1;
}
return roman;
}
}
just change a little bit on the logic
.try on dartpad: https://dartpad.dev/?id
void main() {
print (intToRoman(30)); // result: XXX
}
String intToRoman(int num) {
List<int> numbers = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1];
List<String> romans = ["M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"];
String roman = '';
for (int i = 0; i < numbers.length; i++) {
while (num >= numbers[i]) {
roman += romans[i];
num -= numbers[i];
}
}
return roman;
}
This solution is based on Wiki:
class Solution {
/// digit: 3=Thousands(10³), 2=Hundreds(10²), 1=Tens(10), 0=Units(1)
/// Range for roman numerals: 1...3999
static final romanNumerals = <int,Map<int,String>>{
1 : {3:'M', 2:'C', 1:'X', 0:'I'},
2 : {3:'MM', 2:'CC', 1:'XX', 0:'II'},
3 : {3:'MMM', 2:'CCC', 1:'XXX', 0:'III'},
4 : {2:'CD', 1:'XL', 0:'IV'},
5 : {2:'D', 1:'L', 0:'V'},
6 : {2:'DC', 1:'LX', 0:'VI'},
7 : {2:'DCC', 1:'LXX', 0:'VII'},
8 : {2:'DCCC', 1:'LXXX', 0:'VIII'},
9 : {2:'CM', 1:'XC', 0:'IX'},
};
/* ---------------------------------------------------------------------------- */
Solution();
/* ---------------------------------------------------------------------------- */
String intToRoman(int number) {
if (number < 1 || number >= 4000) return '';
var list = number.toString().split('').map(int.parse).toList();
var buffer = StringBuffer();
final len = list.length;
for (var i = 0; i < len; i++) {
var digit = list[i];
if (digit == 0) continue;
buffer.write(romanNumerals[digit]![len - 1 - i]);
}
return buffer.toString();
}
/* ---------------------------------------------------------------------------- */
void intToRoman2(int number) {
print(intToRoman(number));
}
}
void main(List<String> args) {
Solution()
..intToRoman2(3)
..intToRoman2(58)
..intToRoman2(1994)
;
}
Output:
III
LVIII
MCMXCIV
This code was already sent to LeetCode with the following results:
Runtime: 1130 ms, faster than 27.96% of Dart online submissions for Integer to Roman.
Memory Usage: 150.5 MB, less than 44.09% of Dart online submissions for Integer to Roman.
why not use the simple way?
I use this extension to convert english numbers to persian numbers
extension StringExtensions on String {
String persianNumber() {
String number = this;
number = number.replaceAll("1", "۱");
number = number.replaceAll("2", "۲");
number = number.replaceAll("3", "۳");
number = number.replaceAll("4", "۴");
number = number.replaceAll("5", "۵");
number = number.replaceAll("6", "۶");
number = number.replaceAll("7", "۷");
number = number.replaceAll("8", "۸");
number = number.replaceAll("9", "۹");
number = number.replaceAll("0", "۰");
return number;
}
}
extension IntExtensions on int {
String persianNumber() {
String number = this.toString();
number = number.replaceAll("1", "۱");
number = number.replaceAll("2", "۲");
number = number.replaceAll("3", "۳");
number = number.replaceAll("4", "۴");
number = number.replaceAll("5", "۵");
number = number.replaceAll("6", "۶");
number = number.replaceAll("7", "۷");
number = number.replaceAll("8", "۸");
number = number.replaceAll("9", "۹");
number = number.replaceAll("0", "۰");
return number;
}
}
I have a simple exercise in Coderbyte, it just want to have a function that's WordSplit(strArr) read the array of strings stored in strArr, For example I have two elements like ["hellocat", "apple,bat,cat,goodbye,hello,yellow,why"]
I just want to to determine if the first element in the input can be split into two words, where both words exist in the dictionary that is provided in the second input.
For example: the first element can be split into two words: hello and cat because both of those words are in the dictionary.
So the program should return the two words that exist in the dictionary separated by a comma, as this result hello,cat .
I've made a recursive solution below. It checks if the string to be split starts with any word in the dictionary. If it exists, the function is called again using a substring with that first word removed.
This function only works up to the first word that isn't in the dictionary since you did not specify the expected behavior when the inputted word is not made up of words in the dictionary. You could make it throw an exception perhaps, but please specify your expectation.
void main() {
print(wordSplit(["hellocat", "apple,bat,cat,goodbye,hello,yellow,why"]));
//hello,cat
}
String wordSplit(List<String> arg) {
String wordToSplit = arg[0];
String dict = arg[1];
List<String> parsedDict = arg[1].split(',');
for(String word in parsedDict) {
if(wordToSplit.startsWith(word)) {
//If the substring would be empty, don't do more recursion
if(word.length == wordToSplit.length) {
return word;
}
return word + ',' + wordSplit([wordToSplit.substring(word.length), dict]);
}
}
return wordToSplit;
}
#include <bits/stdc++.h>
using namespace std;
vector<string> converToWords(string dict) {
vector<string> res;
string s = "";
int n = dict.length();
for(int i=0; i<n; i++) {
if(dict[i] == ',') {
res.push_back(s);
s = "";
}
else s += dict[i];
}
res.push_back(s);
s = "";
return res;
}
string solve(string str[]) {
string result = "";
string word = str[0], dict = str[1];
int n = word.length();
vector<string> vs = converToWords(dict);
unordered_set<string> ust;
for(auto it: vs) ust.insert(it);
// for(auto i=ust.begin(); i!=ust.end(); i++){
// cout<<*i<<endl;
// }
string s = "";
for(int i=0; i<n; i++) {
s += word[i];
// cout<<s<<endl;
string temp = word.substr(i+1, n-(i+1));
// cout<<temp<<endl;
if(ust.find(s) != ust.end() && ust.find(temp) != ust.end()) {
cout<<s<<endl;
cout<<temp<<endl;
result += s+","+temp;
break;
}
temp = "";
}
return result;
}
int main() {
string arr[2];
cin>>arr[0]>>arr[1];
cout << solve(arr);
return 0;
}
I'm stuck on this common interview question using Dart. I need to return the most common character in a given string. I'm trying to create a map with a count for each character as the first step.
This is my progress so far:
main(List<String> arguments) {
maxChar('hello');
}
void maxChar(String word) {
Map<String, int> charMap = {};
int max = 0;
String maxChar = '';
word.split('').forEach((char) {
if(charMap.containsValue(char)) {
charMap[char]+1;
return;
} else {
charMap[char] = 1;
}
});
print(charMap);
}
Right now its not even counting the correct amount of the letter 'l'. It's outputting:
{h: 1, e: 1, l: 1, o: 1}
What am I doing wrong? Is there an easier way to return the most common character in a String in Dart?
Thanks!
EDIT:
Ok, I've solved it, but surely there is a more concise way of solving this problem. See my solution below:
main(List<String> arguments) {
print(max_char.maxChar('hello'));
}
String maxChar(String word) {
Map<String, int> charMap = {};
int max = -1;
String maxChar = '';
word.split('').forEach((char) {
if(charMap.containsKey(char)) {
charMap[char]++;
return;
} else {
charMap[char] = 1;
}
});
charMap.forEach((k,v) {
if(v > max) {
max = v;
maxChar = k;
}
});
return maxChar;
}
A shorter approach to counting the characters is definitely possible:
String charCount(String chars) {
int maxChar = -1;
int maxCount = 0;
var counts = <int, int>{};
for (var char in chars.runes) {
int count = counts.update(char, (n) => n + 1, ifAbsent: () => 1);
if (count > maxCount) {
maxCount = count;
maxChar = char;
}
}
return String.fromCharCode(maxChar);
}
If you just want to count the characters, you can remove all the lines mentioning maxCount and maxChar.
I use integers to represent the characters instead of strings. That's cheaper and just as precise, and it allows you to recognize and combine Unicode UTF-16 surrogates.
I would like the optimal solution for removing trailing zeros using Dart. If I have a double that is 12.0 it should output 12. If I have a double that is 12.5 it should output 12.5
I made regular expression pattern for that feature.
double num = 12.50; // 12.5
double num2 = 12.0; // 12
double num3 = 1000; // 1000
RegExp regex = RegExp(r'([.]*0)(?!.*\d)');
String s = num.toString().replaceAll(regex, '');
UPDATE
A better approach, just use this method:
String removeDecimalZeroFormat(double n) {
return n.toStringAsFixed(n.truncateToDouble() == n ? 0 : 1);
}
OLD
This meets the requirements:
double x = 12.0;
double y = 12.5;
print(x.toString().replaceAll(RegExp(r'.0'), ''));
print(y.toString().replaceAll(RegExp(r'.0'), ''));
X Output: 12
Y Output: 12.5
Use NumberFormat:
String formatQuantity(double v) {
if (v == null) return '';
NumberFormat formatter = NumberFormat();
formatter.minimumFractionDigits = 0;
formatter.maximumFractionDigits = 2;
return formatter.format(v);
}
If what you want is to convert a double without decimals to an int but keep it as a double if it has decimals, I use this method:
num doubleWithoutDecimalToInt(double val) {
return val % 1 == 0 ? val.toInt() : val;
}
Lots of the answers don't work for numbers with many decimal points and are centered around monetary values.
To remove all trailing zeros regardless of length:
removeTrailingZeros(String n) {
return n.replaceAll(RegExp(r"([.]*0+)(?!.*\d)"), "");
}
Input: 12.00100003000
Output: 12.00100003
If you only want to remove trailing 0's that come after a decimal point, use this instead:
removeTrailingZerosAndNumberfy(String n) {
if(n.contains('.')){
return double.parse(
n.replaceAll(RegExp(r"([.]*0+)(?!.*\d)"), "") //remove all trailing 0's and extra decimals at end if any
);
}
else{
return double.parse(
n
);
}
}
I found another solution, to use num instead of double. In my case I'm parsing String to num:
void main() {
print(num.parse('50.05').toString()); //prints 50.05
print(num.parse('50.0').toString()); //prints 50
}
Here is what I've come up with:
extension DoubleExtensions on double {
String toStringWithoutTrailingZeros() {
if (this == null) return null;
return truncateToDouble() == this ? toInt().toString() : toString();
}
}
void main() {
group('DoubleExtensions', () {
test("toStringWithoutTrailingZeros's result matches the expected value for a given double",
() async {
// Arrange
final _initialAndExpectedValueMap = <double, String>{
0: '0',
35: '35',
-45: '-45',
100.0: '100',
0.19: '0.19',
18.8: '18.8',
0.20: '0.2',
123.32432400: '123.324324',
-23.400: '-23.4',
null: null
};
_initialAndExpectedValueMap.forEach((key, value) {
final initialValue = key;
final expectedValue = value;
// Act
final actualValue = initialValue.toStringWithoutTrailingZeros();
// Assert
expect(actualValue, expectedValue);
});
});
});
}
String removeTrailingZero(String string) {
if (!string.contains('.')) {
return string;
}
string = string.replaceAll(RegExp(r'0*$'), '');
if (string.endsWith('.')) {
string = string.substring(0, string.length - 1);
}
return string;
}
======= testcase below =======
000 -> 000
1230 -> 1230
123.00 -> 123
123.001 -> 123.001
123.00100 -> 123.001
abc000 -> abc000
abc000.0000 -> abc000
abc000.001 -> abc000.001
Here is a very simple way. Using if else I will check if the number equals its integer or it is a fraction and take action accordingly
num x = 24/2; // returns 12.0
num y = 25/2; // returns 12.5
if (x == x.truncate()) {
// it is true in this case so i will do something like
x = x.toInt();
}
To improve on What #John's answer: here is a shorter version.
String formatNumber(double n) {
return n.toStringAsFixed(0) //removes all trailing numbers after the decimal.
}
This function removes all trailing commas. It also makes it possible to specify a maximum number of digits after the comma.
extension ToString on double {
String toStringWithMaxPrecision({int? maxDigits}) {
if (round() == this) {
return round().toString();
} else {
if (maxDigits== null) {
return toString().replaceAll(RegExp(r'([.]*0)(?!.*\d)'), "");
} else {
return toStringAsFixed(maxDigits)
.replaceAll(RegExp(r'([.]*0)(?!.*\d)'), "");
}
}
}
}
//output without maxDigits:
// 1.0 -> 1
// 1.0000 -> 1
// 0.99990 -> 0.9999
// 0.103 -> 0.103
//
////output with maxDigits of 2:
// 1.0 -> 1
// 1.0000 -> 1
// 0.99990 -> 0.99
// 0.103 -> 0.1
user3044484's version with Dart extension:
extension StringRegEx on String {
String removeTrailingZero() {
if (!this.contains('.')) {
return this;
}
String trimmed = this.replaceAll(RegExp(r'0*$'), '');
if (!trimmed.endsWith('.')) {
return trimmed;
}
return trimmed.substring(0, this.length - 1);
}
}
// The syntax is same as toStringAsFixed but this one removes trailing zeros
// 1st toStringAsFixed() is executed to limit the digits to your liking
// 2nd toString() is executed to remove trailing zeros
extension Ex on double {
String toStringAsFixedNoZero(int n) =>
double.parse(this.toStringAsFixed(n)).toString();
}
// It works in all scenarios. Usage
void main() {
double length1 = 25.001;
double length2 = 25.5487000;
double length3 = 25.10000;
double length4 = 25.0000;
double length5 = 0.9;
print('\nlength1= ' + length1.toStringAsFixedNoZero(3));
print('\nlength2= ' + length2.toStringAsFixedNoZero(3));
print('\nlenght3= ' + length3.toStringAsFixedNoZero(3));
print('\nlenght4= ' + length4.toStringAsFixedNoZero(3));
print('\nlenght5= ' + length5.toStringAsFixedNoZero(0));
}
// output:
// length1= 25.001
// length2= 25.549
// lenght3= 25.1
// lenght4= 25
// lenght5= 1
you can do a simple extension on the double class
and add a function which in my case i called it neglectFractionZero()
in this extension function on double(which returns a string) i
split the converted number to string and i check if the split part of the string is "0" , if so i return the first part only of the split and i neglect this zero
you can modify it according to your needs
extension DoubleExtension on double {
String neglectFractionZero() {
return toString().split(".").last == "0"? toString().split(".").first:toString();
}
}
I've came up with improved version of #John.
static String getDisplayPrice(double price) {
price = price.abs();
final str = price.toStringAsFixed(price.truncateToDouble() == price ? 0 : 2);
if (str == '0') return '0';
if (str.endsWith('.0')) return str.substring(0, str.length - 2);
if (str.endsWith('0')) return str.substring(0, str.length -1);
return str;
}
// 10 -> 10
// 10.0 -> 10
// 10.50 -> 10.5
// 10.05 -> 10.05
// 10.000000000005 -> 10
void main() {
double x1 = 12.0;
double x2 = 12.5;
String s1 = x1.toString().trim();
String s2 = x2.toString().trim();
print('s1 is $s1 and s2 is $s2');
}
try trim method https://api.dartlang.org/stable/2.2.0/dart-core/String/trim.html
Does the compiler create a new location in memory when a record is extended (deep copy?) or does the compiler make the record mutable and modify the value?
For example:
type MyRecord = { A : string
; B : string
}
let record = { A = "A"; B = "B" }
let record = { record with A = "new A" } //copy or overwrite?
Since I am overwriting record does the compiler copy or overwrite? Are there performance concerns either way?
It makes the copy.
Copy-and-update Record expression
*A copy-and-update record expression elaborates as if it were a record expression written as follows:
let v = expr in { field-label1 = expr1 ; … ; field-labeln = exprn; F1 = v.F1; ... ; FM = v.FM }
where F1 ... FM are the fields of R that are not defined in field-initializers and v is a fresh variable.*
This
type T = {
A : string
B : string
}
let x = { A = "a"; B = "b" }
let y = { x with A = "aa" }
is equivalent to this
class T {
public readonly string A;
public readonly string B;
public T(string a, string b) {
A = a;
B = b;
}
}
var x = new T("a", "b");
var y = new T("aa", x.B);