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Editorial Team
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Editorial Team
Asked: May 29, 20262026-05-29T12:30:07+00:00 2026-05-29T12:30:07+00:00

Per the SWIG Documentation (21.9.1 Default primitive type mappings), the C uint8_t is mapped

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Per the SWIG Documentation (21.9.1 Default primitive type mappings), the C uint8_t is mapped to a Java short which is 16 bits and the C uint_15_t is mapped to a Java int which is 32 bits. I believe the C functions are 8 and 16 bits respectively, why does SWIG double the number of bits when wrapping in Java?

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  1. Editorial Team
    Editorial Team
    2026-05-29T12:30:08+00:00Added an answer on May 29, 2026 at 12:30 pm

    The problem is that Java types are always signed.

    Thus if you have an unsigned C type that goes from 0 to 255, the smallest Java type that can represent the upper half of that range is a short.

    The alternative is that you shift or somehow transform your uint8_t to use the negative parts of Java’s byte, but the semantics of that are very counterintuitive.

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