So long back, I was working on a game that added 1 point to the user's score for every correct answer and deducted 0.1 points for a wrong one. The game starts with 1 point in your balance and then the score gets calculated based on your choices. Prima facie it worked fine (kind of) but then something caught me by dismay. I started the game (score = 1) and submitted three wrong answers back to back. What do you expect? A score of 1 - 0.1 - 0.1 - 0.1 = 0.7 ? Got you. Try that right away in your browser's console. It works alright you say? I bet you did 1 - 0.3, that indeed shall give you 0.7 but when you do it incrementally like I did, you shall see that
✔️ 1 - 0.1 = 0.9
✔️ 0.9 - 0.1 = 0.8
❌ 0.8 - 0.1 = 0.7000000000000001
Do we all understand that our computers store all and any kind of information in binary? Assuming you said 'YES', how is a decimal number (which we input) converted into binary before it gets stored? Do you know that after the number is converted into binary, to get stored in the register (memory), the binary should be first arranged in some appropriate format? "Binary Floating Point Representation" is the term we use for those formats. Call it FPR for simplicity.
Binary Floating Point Representation can be of 3 types :
1 - Half Precision Format
- available memory for a given number = 16 bits
- least precise & least wasteful
2 - Single Precision Format
- available memory for a given number = 32 bits
- float data-type in Java
3 - Double Precision Format
- available memory for a given number = 64 bits
- double data-type in Java
- most accurate representation of bigger numbers
Taking you back to school? No, please take a quick look (1.5x speed) at this video if you're not sure what did I just say. Now that you know we have limited space in the memory to store the binary representation, what if the binary of some number you input doesn't fit in 64 bits? Well, we round it up and make it fit in 64 bits somehow and hence we introduce the famous Rounding Error. This rounding error is the characteristic feature of floating-point computation and obviously when you input some number X, it may or may not stay exactly X after binary round off.
So what could be the examples of numbers whose binary won't fit even in 64 bits? A very obvious case can be of a number whose binary representation is non-terminating. 0.1 ? Seriously? Yes, lets see how this simple small decimal number has got a binary equivalent that never terminates (like the value of π in decimal).
There's a simple rule to find out if the given decimal number will have a non terminating binary or not. A decimal has an equivalent terminating binary if and only if the decimal, written as a proper fraction in lowest terms, has a denominator that is a power of two. Example : 0.1 has an infinite binary : 0.1 = 1/10, and 10 is not a power of two. Also 0.5, on the other hand, terminates: 0.5 = 5/10 = 1/2.
1 - Math.round((0.8-0.1)*factor)/factor shall give 0.7, where factor = 10 for rounding to single digit, 100 for rounding the result to 2 digits after decimal and so on.
2 - (0.8-0.1).toFixed(1) shall give "0.7" but in string format. Mostly irrelevant now but "toFixed" may show inconsistencies amongst older versions of some browsers. Read more.
Most decimals have infinite representations in binary. Due to limitation of memory, rounding errors may get introduced in numbers whose binary equivalent does not fit even the Double Precision Format. So do not be surprised the next time you see an anomaly in floating point calculations. Its good to use one of the above mentioned solutions or a custom tailored solution that fits your requirement.