How can I convert exponential numbers to numeric?
For example, I have 4.903381641568837e-7 and want to convert it to 0.00000049034...
You should try the decimal package :
import 'package:decimal/decimal.dart';
void main() {
double smallDouble = 0.0000000000000024353505535353555;
final convertedDouble = Decimal.parse(smallDouble.toString());
print("normal double :>> $smallDouble");
print("Decimal :>> $convertedDouble");
}
Output:
normal double :>> 2.4353505535353554e-15
Decimal :>> 0.0000000000000024353505535353554
Hope it could help.
Related
I hope everything is going well.
I have this unicodestring:
353135313531353135313531
And I want to transform it into another unicodestring with this content:
515151515151
In other words, convert a hex representation into its ASCII interpretation.
It is very straightforward to do this in C, but the idea is to work with C++ Builder.
This is what I have been trying to do:
String hex_to_ascii(const String& hex_str) {
String ascii_str = "";
for (int i = 1; i <= hex_str.Length(); i += 2) {
String hex_char = hex_str.SubString(i, 2);
int ascii_char = hex_char.ToInt();
// ascii_str += String().sprintf(_D("%c"), ascii_char);
ascii_str.Insert(ascii_char, ascii_str.Length() + 1);
}
return ascii_str;
But no luck so far.
I know there is a method called ToHex I've been trying to search for documentation about it because it's related to what I am trying to do, so probably the library that has this method has also something close to what I need.
If you know how to do this or where can I read about the ToHex method, please let me know. Thank you for reading.
The code you have is very close, it just needs some minor tweaks.
Most importantly, String::ToInt() WILL NOT decode hex, like you are expecting. It will convert "35" to an integer with a value of decimal 35 (NOT hex 0x35, decimal 53), and will convert "31" to an integer with a value of decimal 31 (NOT hex 0x31, decimal 49), etc.
You need to instead use Sysutils::StrToInt() with a 0x hex prefix prepended to the string value.
Try this:
String hex_to_ascii(const String& hex_str) {
String ascii_str;
for (int i = 1; i <= hex_str.Length(); i += 2) {
String hex_char = _D("0x") + hex_str.SubString(i, 2);
int ascii_char = StrToInt(hex_char);
ascii_str += static_cast<Char>(ascii_char);
}
return ascii_str;
}
Alternatively, you can use HexToBin() to decode the hex into a byte array, and then construct a UnicodeString from those bytes, eg:
String hex_to_ascii(const String& hex_str) {
TBytes bytes;
bytes.Length = hex_str.Length() / 2;
HexToBin(hex_str.c_str(), &bytes[0], bytes.Length);
return String((char*)&bytes[0], bytes.Length);
// Alternatively:
// return TEncoding::Default.GetString(bytes);
}
i want to map a continuous range of double values [10.0,20.0] to a byte in the range of [100,200];
The following must apply (even when converting back and forth multiple times):
convertToByte(convertToDouble(y)) == y
Thanks.
With help of the comments to the question we're using the following code:
int convertToByte(
double initial,
double minValue,
double maxValue,
int minByte,
int maxByte,
) {
double value = initial - minValue;
double valueRange = maxValue - minValue;
int byteRange = maxByte - minByte;
double valueSteps = valueRange / byteRange;
double byte = (value / valueSteps);
return (minByte + byte.round()).clamp(minByte, maxByte);
}
This does not provide a solution for the specified test, but a deterministic answer for a specific value.
When converting a byte back to a value and vice versa multiple times the output always stays the same.
This is what we needed for our application.
Can't find or work out a solution to this that works with Dart.
I've already tried:
1. toStringAsFixed() - this rounds
2. double.parse(number.toStringAsFixed()) - this rounds
3. num - num % 0.01 - this rounds
I've found some solutions on SO but they either use functions that aren't available on Flutter/Dart or didn't work for me.
Any help would be greatly appreciated.
To build on #Irn beautiful Answer.
The function below lets you specify how many fractional digits / how many decimal places you want
double truncateToDecimalPlaces(num value, int fractionalDigits) => (value * pow(10,
fractionalDigits)).truncate() / pow(10, fractionalDigits);
Example
truncateToDecimalPlace(4321.92385678, 3) // 4321.923
Or my favorite way is to use Dart extensions
extension TruncateDoubles on double {
double truncateToDecimalPlaces(int fractionalDigits) => (this * pow(10,
fractionalDigits)).truncate() / pow(10, fractionalDigits);
}
This lets you call the function on a double object like thus
Usage
4321.92385678.truncateToDecimalPlaces(3); // 4321.923
Try
double truncateToHundreths(num value) => (value * 100).truncate() / 100;
There will be cases where you lose precision, but that requires you to use almost all 53 bits of double precision in the number (when multiplying by 100 loses precision to begin with).
Going trough strings is a significant overhead that I would avoid if possible, unless you actually expect to convert the number to a string anyway afterwards.
(value * 100).truncateToDouble() / 100
Example:
var lat = 29.4562
lat * 100 = 2945.62
2945.62 truncateToDouble() = 2945
2945 / 100 = 29.45
I have used the below method
String toFixed(double value, [int decimalPlace = 1]) {
try {
String originalNumber = value.toString();
List<String> formattedNumber = originalNumber.split('.');
return "${formattedNumber[0]}.${formattedNumber[1].substring(0, decimalPlace)}";
} catch (_) {}
return value.toString();
}
Three part question:
How to find 2 user created horizontal lines on a chart by name and return the price of each.
Then determine which HLine was crossed by the price most recently to determine trend direction.
Calculate Fibonacci levels based on prices and direction
double value = ObjectGetDouble(0,nameOfHLine,OBJPROP_PRICE1);
this is your value if you have name of the object, if you dont have it - need to loop over all objects:
string name;
for(int i=ObjectsTotal()-1;i>=0;i--){
name = ObjectName(i);
if(ObjectType(name)!=OBJ_HLINE) continue;
}
Working example of Fibonacci object that can be edited by the user and printing of fibonacci levels.
#include <ChartObjects/ChartObjectsFibo.mqh>
CChartObjectFibo *Fibo;
int OnInit()
{
Fibo = new CChartObjectFibo();
#Create object and set some defaults
if(!Fibo.Create(0,"Fibonacci",0,Time[5],Open[5],Time[0],Open[0]))
{
return(INIT_FAILED);
}
# Allow user to drag object
Fibo.Selectable(true);
return(INIT_SUCCEEDED);
}
void OnDeinit(const int reason)
{
delete Fibo;
}
void OnTick()
{
string level_description;
double level_value;
string printString="Fibonacci Levels - ";
# Get the two anchor prices
double p1 = Fibo.GetDouble(OBJPROP_PRICE,0);
double p2 = Fibo.GetDouble(OBJPROP_PRICE,1);
# Calculate range
double range=MathAbs(p1-p2);
for(int i=0;i<Fibo.LevelsCount();i++)
{
level_description=Fibo.LevelDescription(i);
# Calculate price of level
level_value=(p2>p1)?p2-range*Fibo.LevelValue(i):p2+range*Fibo.LevelValue(i);
printString=StringFormat("%s %s:%.5f",printString,level_description,level_value);
}
Print(printString);
}
Difficult to understand exactly what you are after, not sure if you are trying to find the graphical objects or just calculate levels based on the prices. Assuming you have the price of the two horizontal lines, the following structure and function can be used to calculate Fibonacci levels. (price 1 is earlier in time than price 2).
Calculation based on formula found here
struct FibLevel {
double retrace38;
double retrace50;
double retrace61;
double extension61;
double extension100;
double extension138;
double extension161;
};
void FibLevel(double price1, double price2,FibLevel &fiblevel)
{
double range = MathAbs(price1-price2);
fiblevel.retrace38 =(price1<price2)?price2-range*0.382:price1+range*0.382;
fiblevel.retrace50 =(price1<price2)?price2-range*0.500:price1+range*0.500;
fiblevel.retrace61 =(price1<price2)?price2-range*0.618:price1+range*0.618;
fiblevel.extension61 =(price1<price2)?price2+range*0.618:price1-range*0.618;
fiblevel.extension100=(price1<price2)?price2+range :price1-range;
fiblevel.extension138=(price1<price2)?price2+range*1.382:price1-range*1.382;
fiblevel.extension161=(price1<price2)?price2+range*1.618:price1-range*1.618;
}
I'm trying to create a swift iOS program that converts a number into dec, bin, and hex numbers. I've come across the strtoul function, but don't quite understand how to use it, would someone be able to explain it? Thanks!
The method strtoul is pretty simple to use. You will need also to use String(radix:()) to convert it to the other direction. You can create an extension to convert from hexaToDecimal or from binaryToDecimal as follow:
Usage String(radix:())
extension Int {
var toBinary: String {
return String(self, radix: 2)
}
var toHexa: String {
return String(self, radix: 16)
}
}
Usage strtoul()
extension String {
var hexaToDecimal: Int {
return Int(strtoul(self, nil, 16))
}
var hexaToBinary: String {
return hexaToDecimal.toBinary
}
var binaryToDecimal: Int {
return Int(strtoul(self, nil, 2))
}
var binaryToHexa: String {
return binaryToDecimal.toHexa
}
}
Testing
let myBinFromInt = 255.toBinary // "11111111"
let myhexaFromInt = 255.toHexa // "ff"
let myIntFromHexa = "ff".hexaToDecimal // 255
let myBinFromHexa = "ff".hexaToBinary // "11111111"
let myIntFromBin = "11111111".binaryToDecimal // 255
let myHexaFromBin = "11111111".binaryToHexa // "ff"
The strtoul() function converts the string in str to an unsigned long
value. The conversion is done according to the given base, which must be between 2 and 36 inclusive, or be the special value 0.
Really it sounds like you want to use NSString
From what it sounds like, you want to convert an unsigned integer to decimal, hex and binary.
For example, if you had an integer n:
var st = NSString(format:"%2X", n)
would convert the integer to hexadecimal and store it in the variable st.
//NSString(format:"%2X", 10) would give you 'A' as 10 is A in hex
//NSString(format:"%2X", 17) would give you 11 as 17 is 11 in hex
Binary:
var st = NSString(format:"%u", n)
Decimal (2 decimal places)
var st = NSString(format:"%.02f", n)