float vs double（在Java中）

Question

Is there ever a case where these two methods would return different values given the same inputs?

int compare1(float a, float b)
{
    return Double.compare(a, b);
}

int compare2(float a, float b)
{
    return Float.compare(a, b);
}

By the same token, is it true (or false) that any number storable in a java's Float can be stored in a java's Double without losing any precision?

Thanks

Answer 1

Yes; casting doubles to floats can yield different results.

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If the difference between a and b is too small to show up in a float, compare2() will return 0 whereas compare1() would not.

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You just edited the question to reverse what you were asking. The new answer is:

I'm almost certain that they will always be the same.

Answer 2

Every float can be represented exactly as a double . From this it follows that the two functions always return the same result.

Answer 3

A double just gives you more bits of precision beyond the decimal place than the float . If all those those extra bits are zero then you have the same value as the float. So yes, all float s can be converted to double s without losing precision.