Code:

class Main {

public static void main (String[] args) {

System.out.print("float: ");

System.out.println(1.35f-0.00026f);

System.out.print("double: ");

System.out.println(1.35-0.00026);

}

}

Output:

float: 1.34974

double: 1.3497400000000002

??? float got the right answer, but double is adding extra stuff from no where, Why??

Isn't double supposed to be more precise than float?

解決方案

A float is 4 bytes wide, whereas a double is 8 bytes wide.

Surely the double has more precision so it has slightly less rounding error.

Squeezing infinitely many real numbers into a finite number of bits

requires an approximate representation. Although there are infinitely

many integers, in most programs the result of integer computations can

be stored in 32 bits. In contrast, given any fixed number of bits,

most calculations with real numbers will produce quantities that

cannot be exactly represented using that many bits. Therefore the

result of a floating-point calculation must often be rounded in order

to fit back into its finite representation. This rounding error is the

characteristic feature of floating-point computation.

On a side note:-

I would suggest if you want the exact decimal values then use java.math.BigDecimal

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