This is fine, remember that using the constructor of BigDecimal to declare a value can be dangerous when it's not of type String. Consider the below...
BigDecimal valDouble = new BigDecimal(0.35);
System.out.println(valDouble);
This will not print 0.35, it will infact be...
0.34999999999999997779553950749686919152736663818359375
I'd say your solution is probably the safest because of that.
Answer from David on Stack OverflowThis is fine, remember that using the constructor of BigDecimal to declare a value can be dangerous when it's not of type String. Consider the below...
BigDecimal valDouble = new BigDecimal(0.35);
System.out.println(valDouble);
This will not print 0.35, it will infact be...
0.34999999999999997779553950749686919152736663818359375
I'd say your solution is probably the safest because of that.
Can we lose precision with toString() method ?
Kind of ... Both Float.toString() and Double.toString() only output the number of digits after the decimal separator, which is required for the output uniquely to correspond to a float or double value.
To use the 0.35 example in david99world's answer, consider the following code:
BigDecimal bd1 = new BigDecimal(0.35);
Number n = 0.35;
BigDecimal bd2 = new BigDecimal(n.toString());
System.out.println(bd1);
System.out.println(bd2);
An intuitive expectation may be that the two BigDecimal instances are identical, but the output shows that they are not:
0.34999999999999997779553950749686919152736663818359375
0.35
The first line is the exact value of the double, since 0.35 cannot be represented exactly. The second line is 0.35, since no more fractional digits are required to represent the distinct value. E.g. the statement 0.34999999999999997779553950749686919152736663818359375 == 0.35 will evaluate to true.
This is actually not a loss of precision when creating the BigDecimal, the uncertainty is already there in your "source" value. The problem is rather that the discrete values possible using e.g. a float or double value as source not necessarily will be represented by the exact equivalent in the BigDecimal instance.
To multiply an integer (or byte/short/float/double) with a BigInteger (or BigDecimal), you must convert the native number to BigInteger/BigDecimal first.
// int parameter can be int or Integer
public static BigInteger multiply ( int a, BigInteger b ) {
return BigInteger.valueOf( a ).multiply( b );
}
// BigInteger <> BigDecimal
public static BigDecimal multiply ( int a, BigDecimal b ) {
return BigDecimal.valueOf( a ).multiply( b );
}
// same for add, subtract, divide, mod etc.
Note:
valueOfis not the same asnew, and for different reasons on BigDecimal and BigInteger. In both cases, I recommendvalueOfovernew.
I see that you added your code, nice.
It doesn't work because Integer is mixed with BigDecimal, and also * does not work with BigDecimal.
If you compare it with my code, the fix should be obvious:
public BigDecimal methCal ( int quantite, BigDecimal prixUnit ) {
return BigDecimal.valueOf( quantite ).multiply( prixUnit );
}
Google definitely could have helped you, if you know what to look for:
https://docs.oracle.com/javase/9/docs/api/java/math/BigDecimal.html#BigDecimal-int-
This is one of the constructors for BigDecimal, which allows you to do the following:
BigDecimal five = BigDecimal.valueOf(5);
BigDecimal seven = BigDecimal.valueOf(2).add(five);
Seeing as you stated you wanted to multiply an int and a BigDecimal, this would be achieved as follows:
BigDecimal result = yourBigDecimal.multiply(BigDecimal.valueOf(yourInt));
And, supposing you want this result as an int:
int intResult = result.intValue();
Keep in mind that this throws away the fraction though. If you want rounding instead:
int intResult = result.round(0, RoundingMode.HALF_UP).intValue();
The BigDecimal(double) constructor can have unpredictable behaviors. It is preferable to use BigDecimal(String) or BigDecimal.valueOf(double).
Use with AISystem.out.println(new BigDecimal(135.69)); //135.68999999999999772626324556767940521240234375
System.out.println(new BigDecimal("135.69")); // 135.69
System.out.println(new BigDecimal("13۵.۶9")); // 135.69
System.out.println(BigDecimal.valueOf(135.69)); // 135.69
The documentation for BigDecimal(double) explains in detail:
- The results of this constructor can be somewhat unpredictable. One might assume that writing new BigDecimal(0.1) in Java creates a BigDecimal which is exactly equal to 0.1 (an unscaled value of 1, with a scale of 1), but it is actually equal to 0.1000000000000000055511151231257827021181583404541015625. This is because 0.1 cannot be represented exactly as a double (or, for that matter, as a binary fraction of any finite length). Thus, the value that is being passed in to the constructor is not exactly equal to 0.1, appearances notwithstanding.
- The String constructor, on the other hand, is perfectly predictable: writing new BigDecimal("0.1") creates a BigDecimal which is exactly equal to 0.1, as one would expect. Therefore, it is generally recommended that the String constructor be used in preference to this one.
- When a double must be used as a source for a BigDecimal, note that this constructor provides an exact conversion; it does not give the same result as converting the double to a String using the Double.toString(double) method and then using the BigDecimal(String) constructor. To get that result, use the static valueOf(double) method.
Use with AIString currency = "135.69";
System.out.println(new BigDecimal(currency));
//will print 135.69