How to simplify the price normalisation function - General

Slava 2016.09.27 17:33 #151

fxsaber:
Obsolete

The standard library has a granularity-aware price normalisation function

//+------------------------------------------------------------------+
//| Normalize price                                                  |
//+------------------------------------------------------------------+
double CSymbolInfo::NormalizePrice(const double price) const
  {
   if(m_tick_size!=0)
      return(NormalizeDouble(MathRound(price/m_tick_size)*m_tick_size,m_digits));
//---
   return(NormalizeDouble(price,m_digits));
  }

fxsaber 2016.09.27 17:54 #152

Slawa:

The standard library has a granularity-aware price normalisation function

I am aware of this extremely slow implementation. It was about fine-tuning NormalizeDouble (and Help) to the current realities. Everyone can write their own granularities.

Sergei Vladimirov 2016.09.27 18:07 #153

transcendreamer:
I was beginning to guess that NormalizeDouble(new_lot-sum_lots,Lots_Digits); it doesn't output exactly 0 and stores some tail

If variables new_lot and sum_lots are equal, then the difference is exactly 0. But it is unknown how you calculate them, they may indeed be unequal when calculated, hence the non-zero difference. Do the same thing like this:

NormalizeDouble(new_lot, Lots_Digits) - NormalizeDouble(sum_lots, Lots_Digits).

If variables are equal within specified number of digits, then the difference will be strictly 0.

Errors, bugs, questions Combining two expert advisors Why is Open[0] the

חולםטרנסצנדר ᨖ 2016.09.27 19:07 #154

Sergei Vladimirov:

If variables new_lot and sum_lots are equal, then the difference is exactly 0. But it is unknown how you calculate them, they may indeed be unequal when calculated, hence the non-zero difference. Do the same thing like this:

NormalizeDouble(new_lot, Lots_Digits) - NormalizeDouble(sum_lots, Lots_Digits).

If the variables are equal within the specified number of digits, the difference will be strictly 0.

oh, thank you!

Sergei Vladimirov 2016.09.27 19:26 #155

transcendreamer:

Again about rounding......

Please advise on the situation (don't throw tomatoes, I'm a humanitarian),

there is such a variable:

double delta=NormalizeDouble(new_lot-sum_lots,Lots_Digits);

if(delta>0) delta-=OrderLots();

if(delta<0) delta+=OrderLots();

The delta is originally normalized,

OrderLots should probably return normalized dubs,

but somehow sometimes on rare occasions I get numbers like 2.775557561562891e-17

So it's almost zero but not zero.......

first question - is this normal?

...

I read it again carefully. It's not normal. If NormalizeDouble function operation :

The integer part is selected - I
Fractional part is selected - F
F = F * 10^digits
F = F (+ or - depending on the sign) 0.5
F = (integer part of F) / 10^digits
result = I + F

then also theNormalizeDouble(new_lot - sum_lots, Lots_Digits) resultshould be strictly zero if new_lot and sum_lots are equal within the specified number of digits. In rare cases there may be a difference of 1 in the last digit (the reason is in items 4 and 5), but the result 2.775557561562891e-17 is strange, it should not be.

How do I go [ARCHIVE] Any rookie question, Any questions from newcomers

pavlick_ 2016.09.27 22:14 #156

I wrote a little tutorial code (I was interested in poking around myself) that turns out the innards of a floating number. If anyone is interested, you can run it (C++ code, you can use some online compiler. Here https://goo.gl/tP691X, for example)

#include <iostream>
using namespace std;

int main()
{
    cout.precision(17);
    float f = 0.5;
    //float add = 0.00000002980232239; //-25   1/(2^25)
    float add = 0.00000005960464478; //-24     1/(2^24)
    //float add = 0.0000001192092896; //-23   1/(2^23)
    //float add = 0.0000002384185791; // -22  1/(2^22)
    f+= add;

    cout << "value = " << f << endl;
    cout << "----------\n";

    char *c = (char*)&f;  // char может ссылаться на любой тип
    cout << "mantissa:\n";
    cout << "implicit_1 ";
    cout << (c[2] >> 6 & 0 b1) << ' '; cout << (c[2] >> 5 & 0 b1) << ' ';
    cout << (c[2] >> 4 & 0 b1) << ' '; cout << (c[2] >> 3 & 0 b1) << ' ';
    cout << (c[2] >> 2 & 0 b1) << ' '; cout << (c[2] >> 1 & 0 b1) << ' ';
    cout << (c[2] >> 0 & 0 b1) << ' '; cout << (c[1] >> 7 & 0 b1) << ' ';
    cout << (c[1] >> 6 & 0 b1) << ' '; cout << (c[1] >> 5 & 0 b1) << ' ';
    cout << (c[1] >> 4 & 0 b1) << ' '; cout << (c[1] >> 3 & 0 b1) << ' ';
    cout << (c[1] >> 2 & 0 b1) << ' '; cout << (c[1] >> 1 & 0 b1) << ' ';
    cout << (c[1] >> 0 & 0 b1) << ' '; cout << (c[0] >> 7 & 0 b1) << ' ';
    cout << (c[0] >> 6 & 0 b1) << ' '; cout << (c[0] >> 5 & 0 b1) << ' ';
    cout << (c[0] >> 4 & 0 b1) << ' '; cout << (c[0] >> 3 & 0 b1) << ' ';
    cout << (c[0] >> 2 & 0 b1) << ' '; cout << (c[0] >> 1 & 0 b1) << ' ';
    cout << (c[0] >> 0 & 0 b1) << '\n';
    cout << "exponenta E - 127:\n";
    cout << "1= " << (c[2] >> 7 & 0 b1) << '\n';
    cout << "2= " << (c[3] >> 0 & 0 b1) << '\n';
    cout << "4= " << (c[3] >> 1 & 0 b1) << '\n';
    cout << "8= " << (c[3] >> 2 & 0 b1) << '\n';
    cout << "16= " << (c[3] >> 3 & 0 b1) << '\n';
    cout << "32= " << (c[3] >> 4 & 0 b1) << '\n';
    cout << "64= " << (c[3] >> 5 & 0 b1) << '\n';
    cout << "128= " << (c[3] >> 6 & 0 b1) << '\n';
    cout << "sign\n";
    cout << (c[3] >> 7 & 0 b00000001) << '\n';
}

The output at f == 0.5 + 1/(2^24). 1/(2^24) is the least significant digit of the mantissa at a given degree:

value = 0.50000005960464478
----------
mantissa:
implicit_1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
exponenta E - 127:
1= 0
2= 1
4= 1
8= 1
16= 1
32= 1
64= 1
128= 0
sign
0

Degree = 126 - 127 = -1

mantissa = 1.0000000000000000000000001

Shift the mantissa to degree -1 = 0.1000000000000000000001 = 1^(-1) + 1^(-24) = 1/(2^1) + 1/(2^24) = 0.5 + 0.00000005960464478 = 0.50000005960464478

As a theory https://habrahabr.ru/post/112953/.

SZZ: this online compiler is nicer than http://rextester.com/l/cpp_online_compiler_gcc

Questions from a "dummy" Time to convert libraries Error opening the file

fxsaber 2016.09.27 22:30 #157

pavlick_:

I wrote a little tutorial code (I was interested in poking around myself) that turns out the innards of a floating number. If anyone is interested, you can run it (C++ code, you can use some online compiler. Here https://www.tutorialspoint.com/compile_cpp11_online.php, for example)

You can also run it in MQL5 - replace c[Pos] with _R(f)[(char)Pos] (or _R(f).Bytes[Pos]) by connecting this library.

SZ

#include "fmtprntl.mqh"    // https://www.mql5.com/en/blogs/post/680570
#include <TypeToBytes.mqh> // https://www.mql5.com/ru/code/16280

#define  c _R(f).Bytes
#define  endl "\n"

void OnStart()
{
  OutputStream cout(16, ' ');
  
  float f = 0.5;
  //float add = 0.00000002980232239; //-25   1/(2^25)
  float add = 0.00000005960464478; //-24     1/(2^24)
  //float add = 0.0000001192092896; //-23   1/(2^23)
  //float add = 0.0000002384185791; // -22  1/(2^22)
  f+= add;

  cout << "value = " << f << endl;
  cout << "----------\n";

  cout << "mantissa:\n";
  cout << "implicit_1 ";
  cout << (c[2] >> 6 & 1) << ' '; cout << (c[2] >> 5 & 1) << ' ';
  cout << (c[2] >> 4 & 1) << ' '; cout << (c[2] >> 3 & 1) << ' ';
  cout << (c[2] >> 2 & 1) << ' '; cout << (c[2] >> 1 & 1) << ' ';
  cout << (c[2] >> 0 & 1) << ' '; cout << (c[1] >> 7 & 1) << ' ';
  cout << (c[1] >> 6 & 1) << ' '; cout << (c[1] >> 5 & 1) << ' ';
  cout << (c[1] >> 4 & 1) << ' '; cout << (c[1] >> 3 & 1) << ' ';
  cout << (c[1] >> 2 & 1) << ' '; cout << (c[1] >> 1 & 1) << ' ';
  cout << (c[1] >> 0 & 1) << ' '; cout << (c[0] >> 7 & 1) << ' ';
  cout << (c[0] >> 6 & 1) << ' '; cout << (c[0] >> 5 & 1) << ' ';
  cout << (c[0] >> 4 & 1) << ' '; cout << (c[0] >> 3 & 1) << ' ';
  cout << (c[0] >> 2 & 1) << ' '; cout << (c[0] >> 1 & 1) << ' ';
  cout << (c[0] >> 0 & 1) << '\n';
  cout << "exponenta E - 127:\n";
  cout << "1= " << (c[2] >> 7 & 1) << '\n';
  cout << "2= " << (c[3] >> 0 & 1) << '\n';
  cout << "4= " << (c[3] >> 1 & 1) << '\n';
  cout << "8= " << (c[3] >> 2 & 1) << '\n';
  cout << "16= " << (c[3] >> 3 & 1) << '\n';
  cout << "32= " << (c[3] >> 4 & 1) << '\n';
  cout << "64= " << (c[3] >> 5 & 1) << '\n';
  cout << "128= " << (c[3] >> 6 & 1) << '\n';
  cout << "sign\n";
  cout << (c[3] >> 7 & 1) << '\n';
}

Result

2016.09.27 23:56:01.178 Test (Si-12.16,H1)       0
2016.09.27 23:56:01.178 Test (Si-12.16,H1)      sign
2016.09.27 23:56:01.178 Test (Si-12.16,H1)      128=  0
2016.09.27 23:56:01.178 Test (Si-12.16,H1)      64=  1
2016.09.27 23:56:01.178 Test (Si-12.16,H1)      32=  1
2016.09.27 23:56:01.178 Test (Si-12.16,H1)      16=  1
2016.09.27 23:56:01.178 Test (Si-12.16,H1)      8=  1
2016.09.27 23:56:01.178 Test (Si-12.16,H1)      4=  1
2016.09.27 23:56:01.178 Test (Si-12.16,H1)      2=  1
2016.09.27 23:56:01.178 Test (Si-12.16,H1)       1=  0
2016.09.27 23:56:01.178 Test (Si-12.16,H1)      exponenta E - 127:
2016.09.27 23:56:01.178 Test (Si-12.16,H1)       implicit_1  0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 1
2016.09.27 23:56:01.178 Test (Si-12.16,H1)       mantissa:
2016.09.27 23:56:01.178 Test (Si-12.16,H1)       ----------
2016.09.27 23:56:01.178 Test (Si-12.16,H1)      value =  0.5000000596046448

pavlick_ 2016.09.27 23:15 #158

implicit_1  0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 0 32 1

What happened to the mantissa?

32 is the ascii code of the problem. It looks like the library with the error does not have a char version. I think we need to remove the problems between mantissa values. Or maybe instead of ' ' write " "?

Roffild's library Multi Timeframe Indicators [WARNING CLOSED!] Any newbie

fxsaber 2016.09.27 23:31 #159

pavlick_:

What happened to the mantissa?

32 is the ascii code of the problem. It looks like a library with a bug. I think we need to remove the problems between mantissa values. Or maybe instead of ' ' write " "?

The fmtprntl.mqh is not mine, so I can't say for sure.

I don't want to bother, so for your f printed byte values

2016.09.28 00:34:09.813 Test (Si-12.16,H1)      c[0] = 1
2016.09.28 00:34:09.813 Test (Si-12.16,H1)      c[1] = 0
2016.09.28 00:34:09.813 Test (Si-12.16,H1)      c[2] = 0
2016.09.28 00:34:09.813 Test (Si-12.16,H1)      c[3] = 63

Alexander Bereznyak 2016.09.28 00:03 #160

Slawa:

A related side effect.

It turned out to be convenient. But it was not originally intended to be used in this way.

There are special functions for printing real numbers with the required accuracy.

Tell me why you need to round real numbers in the calculation process? Because in this case, you lose calculation accuracy!

We are talking about the correct comparison of prices, stops, lots

You need a certain accuracy here.

Rounding numbers in MT4 via NormalizeDouble - page 16