This question already has answers here :
A float is a binary representation of a number ( 1 / 2 ^^ ? )
like:
1/2^1 = 0.5,
1/2^2 = 0.25,
1/2^3 = 0.125,
1/2^4 = 0.0625, ...
The result renders from an addition of this binary numbers to find the closest match.
Printing float values with fixed number of decimal places through , How to print float number (value) with fixed number of decimal place using cout in C++ language? C++ program to Read more: std::setprecision. Consider the� In this output the number of digits after decimal are 6, which is default format of float value printing. Now, we want to print 2 digits only after decimal. Use "%.nf" format specifier. By using this format specifier we can print specific number of digits after the decimal, here "n" is the number of digits after decimal point.
Never use floats or doubles for prices, floats and doubles are approximations, see: IEEE_754 .
Have a look at BigDecimal instead.
Edit to be exact: use floats or doubles to represents decimals is an approximation.
5 Examples of Formatting Float or Double Numbers to String in Java , This will format the floating point number 1.23456 up-to 2 decimal places, way, even DecimalFormat rounds the number if next decimal point is more than 5. Note: If the number in the third decimal place is more than 5, the 2nd decimal place value increases to 1 in the output. How to Round Float to 3 Decimal Places in Python. If you want to round the float value to 3 decimal places. You have to use the same round() and pass 3 as the second argument. The first argument is the same float variable you
That happens because floating point values are often imprecise for decimal fractions .
Basically, since floats are represented using negative powers of 2, some decimal values are very hard to represent accurately (e.g.: how do you write 0.3 as a sum of powers of two?).
You can try Float.valueOf("1602.720000") or Double.valueOf("1602.72") (better), but those will probably have the same issues.
Refer to this answer for a more detailed explanation.
How to print a float with two decimal places in Python, Use str. format() to print a float with two decimal places format(number) with "{:. 2f}" as str and a float as number to return a string representation of the number with two decimal places. Call print(string) with the formatted float-string as string to print the float. We can print float numbers with fixed number of decimal places using std::fixed and std::setprecision, these are the manipulators, which are defined in the iomanip header file. Syntax of setprecision. std::setprecision(int n) Here, n is the number of digits after the decimal point (number of decimal places) Read more: std::setprecision
GENERAL: PRECISION OF PRINTF %F FORMAT STRING, SYMPTOMS: I understand that the single precision floating-point values only have a sense to give users the impression that a floating-point number has more precision. Therefore, the printf routine implements rounding after 7 decimal digits. I am experiencing a very strange behaviour when i am trying to print float number with more decimal points. Float price = Float.valueOf("1602.72"); System.out.println(price); //prints 1602.72
Only 2 decimal places in printed float, And the serial monitor showed 1.45 as an output. How can increase the precision of the float I am saving in memory? share. How to print floating point numbers with a specified precision? Rounding is not required. For example, 5.48958123 should be printed as 5.4895 if given precision is 4. For example, below program sets the precision for 4 digits after the decimal point:
How to display a float with two decimal places in Python?, You can use string formatting to format floating point numbers to a fixed width in Python.For example, if you want the decimal points to be� How to display a float with two decimal places in Python? Python Server Side Programming Programming You can use string formatting to format floating point numbers to a fixed width in Python.For example, if you want the decimal points to be aligned with width of 12 characters and 2 digits on the right of the decimal, you can use the following:
Comments So, there is no method to achieve trailing zeros? Closely related to this even if it is tagged specifically for android: stackoverflow.com/q/35673034/4636715 So, consider using BigDecimal So there is no method to do it in float and decimal. I am now appending 0's to string. Thanks A float is not a binary representation of a decimal number. A float is a binary representation of a number. All numerical arithmetic at best approximates real arithmetic. It is incorrect to suggest that BigDecimal is not an approximation while binary floating-point is. For example, BigDecimal cannot represent one-third. Correct statements of the differences are that BigDecimal uses a decimal base, and hence can exactly represent decimal numerals, and that BigDecimal provides arbitrary precision rather than IEEE-754’s fixed precision.BigDecimal represents decimal numbers, not real one, so it's not an approximation. The question is about price which is also decimal, so we can exactly represent price with BigDecimal. How much is this sandwich? 6π$ :D So, yes representing real with BigDecimal will be an approximation, but my answer is context oriented, natural language is context oriented… If BigDecimal is not on approximation because it represents decimal numbers, then float and double are not approximations because they represent binary numbers. BigDecimal is subject to errors the same as binary floating-point is: As noted, dividing 1 by 3 in BigDecimals produces result that is not mathematically correct. That means, if a merchant offers a buy-two-get-one-free deal, BigDecimal cannot represent the effective unit price. Ok, but the author of the question wants to represent price! So, decimals. As noted with my example, prices are not always in decimal. Nor does decimal automatically solve all problems in currency calculations. It is misleading to suggest that binary floating-point arithmetic is merely an approximation while BigDecimal is not. They are both approximations, but they are different approximations. Since your answer suggests one is an approximation, and the other is not, it is misleading. They're not "very hard," they're impossible. :-)
