**Pedro Taveira:**I'm runnng this code to increase the lotsize in steps:

What I get is this result:

Why is not the result of the calculation a value with 2 decimal cases?

I'm just adding two numbers with 2 decimal cases each.

Forum on trading, automated trading systems and testing trading strategies

MathRound fails for one particular number

Fernando Carreiro, 2018.01.01 22:08

He means that the value "0.69" cannot be exactly represented given the way a floating point number works (based on binary and not decimal representation) - hence, why the value gets represented as "0.68999999..." (see below).

You can never really "normalize" it and it is the main reason why both @whroeder1 and myself contest the use of the *NormalizeDouble()* function. It should in the very least be renamed to something like *"RoundDigits()"* because it is more of a "rounding" function and does not "normalize" in any way or fashion.

https://www.h-schmidt.net/FloatConverter/IEEE754.html

**EDIT: Please note that the above images are for examples of representation in the 4-byte "float", and not the 8-byte "double" which offers more precision but still cannot represent the value "0.69" exactly due to the "binary" nature of the format.**

EDIT2: For future readers that have difficulty understanding (or accepting) this concept, of decimal values not having an exact representation with a "float" or a "double", here is an article worth reading:

# Why 0.1 Does Not Exist In Floating-Point

Many new programmers become aware of binary floating-point after seeing their programs give odd results: “Why does my program print 0.10000000000000001 when I enter 0.1?”; “Why does 0.3 + 0.6 = 0.89999999999999991?”; “Why does 6 * 0.1 not equal 0.6?” Questions like these are asked every day, on online forums like stackoverflow.com.

The answer is that most decimals have infinite representations in binary. Take 0.1 for example. It’s one of the simplest decimals you can think of, and yet it looks so complicated in binary:

Decimal 0.1 In Binary ( To 1369 Places

The bits go on forever; no matter how many of those bits you store in a computer, you will never end up with the binary equivalent of decimal 0.1.

... Read the rest of the article at: http://www.exploringbinary.com/why-0-point-1-does-not-exist-in-floating-point/

I'm runnng this code to increase the lotsize in steps:

What I get is this result:

Why is not the result of the calculation a value with 2 decimal cases?

I'm just adding two numbers with 2 decimal cases each.

Thanks.

Pedro