Round to the low

New comment
 

Hi all, could you help me with this problem?

int start() 
  { 
          
Print(NormalizeDouble(0.09/2,2));

  }

In this case output is 0.05.

I need to obtain 0.04. How can I solve?

Thank you!

 
Alberto_jazz:

Hi all, could you help me with this problem?

In this case output is 0.05.

I need to obtain 0.04. How can I solve?

Thank you!

If you explain why you are trying to do what you asked you may get a better answer. Otherwise, subtract 0.001 before normalizing.
 
Alberto_jazz:
Print(NormalizeDouble(0.09/2,2));
In this case output is 0.05.
I need to obtain 0.04.
  1. since 0.09/2 is 0.045, rounding to 2 decimals gives 0.05
  2. If you need to truncate try
    double TruncateDouble(double v, int d){
   double mulip = MathPow(10.0,d),
          trunc = MathFloor(v * multi) / multp;
   return(trunc);   
}
/////////////
Print(TruncateDouble(0.09/2,2)); // 0.04
  3. You should change the title of your post "Round to the low" to "Truncate to the low"
 

@WHRoeder

double TruncateDouble(double v, int d){
//   double mulip = MathPow(10.0,d),
//          trunc = MathFloor(v * multi) / multp;
   double mulip = MathPow(10.0,d),
          trunc = MathFloor(v * mulip) / mulip;
   return(trunc);   
}

since I am totally mathematically challenged your quickie was totally what needed. I was doing eg,

int i=0.498553*100;
double d=i/100.0;
Print(d);  //0.49

function is now resident in my tools mqh kit bag!

took liberty to c/mul*/mulip/all

was stumped to begin with as totally out of my depth. was do to *ip,*ti,*tp since this is all .ru to me ;-) but got there in the end. 

 thank you 

New comment