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Home/ Questions/Q 7762815
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T14:30:18+00:00 2026-06-01T14:30:18+00:00

If I’m working with a double , and I convert it to a float

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If I’m working with a double, and I convert it to a float, how does this work exactly? Does the value get truncated so it fits into a float? Or does the value get rounded differently? Sorry if this sounds a bit remedial, but I’m trying to grasp the concept of float and double conversions.

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  1. Editorial Team
    Editorial Team
    2026-06-01T14:30:19+00:00Added an answer on June 1, 2026 at 2:30 pm

    From the Java Language Specification, section 5.1.3:

    A narrowing primitive conversion from double to float is governed by the IEEE 754 rounding rules (§4.2.4). This conversion can lose precision, but also lose range, resulting in a float zero from a nonzero double and a float infinity from a finite double. A double NaN is converted to a float NaN and a double infinity is converted to the same-signed float infinity.

    and section 4.2.4 says:

    The Java programming language requires that floating-point arithmetic behave as if every floating-point operator rounded its floating-point result to the result precision. Inexact results must be rounded to the representable value nearest to the infinitely precise result; if the two nearest representable values are equally near, the one with its least significant bit zero is chosen. This is the IEEE 754 standard’s default rounding mode known as round to nearest.

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