Conversion from int16_t to uint32_t

桃扇骨 2022-05-25 07:24 192阅读 0赞

int16_t s; uint32_t ui = s;

Is the result of converting a `int16_t` value to a `uint32_t` value compiler dependent? If not, what's the rule?

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The results are well-defined; non-negative values stay the same, and negative values are reduced modulo 2^32. But the situations where exact sized types like `int16_t` and `uint32_t` are quite rare. There's really no need for anything other than `int` and `unsigned long` here: those types have at least as many bits as `int16_t`and `uint32_t` and, unlike `int16_t` and `uint32_t`, they're required to exist on any conforming implementation. If you really really want the sexy new sized types, at least go for portability with`int_least16_t` and `uint_least32_t`.

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look for "integer conversions" in the standard. The difference between `uint32_t` and`uint_least32_t` here is precisely that the value of a converted `int16_t` is fully determined by the standard with `uint32_t`, whereas with `uint_least32_t` it depends on the width of the type (and only on that, since once the implementation decides that the rest of the behavior is determined). In the unlikely event that you want to convert negative values to an unsigned type, chances are you do want to know the width of the type. – [Steve Jessop][]

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[Steve Jessop]: http://stackoverflow.com/users/13005/steve-jessop