Bit Fields

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Originally Posted by Zaccheus

SuperKoko ... does that not simply mean that a 16 bit un/signed bitfield member will behave the same as a 16 bit un/signed short ?

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For signed bit-fields, this is implementation-dependent.
Indeed, conversions from integer types to signed bit-fields give implementation-defined results, or raise an implementation-defined signal if the value can't be preserved.
On the other hand, integer overflow has undefined behavior.

A 16 bits unsigned bit-field will behave like a 16 bits unsigned integer (i.e. an unsigned integer having 16 value bits and 0 or more padding bits).

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Originally Posted by Plasmator

So I can use this okay, right?

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You can use it safely for unsigned bit-fields.

For signed bit-fields, this isn't a great idea since results will be implementation-dependent.

And, if you fear portability issues, you can use the % operator.

Quote:

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Originally Posted by SuperKoko

For signed bit-fields, this is implementation-dependent.
Indeed, conversions from integer types to signed bit-fields give implementation-defined results, or raise an implementation-defined signal if the value can't be preserved.
On the other hand, integer overflow has undefined behavior.

A 16 bits unsigned bit-field will behave like a 16 bits unsigned integer (i.e. an unsigned integer having 16 value bits and 0 or more padding bits).


You can use it safely for unsigned bit-fields.

For signed bit-fields, this isn't a great idea since results will be implementation-dependent.

And, if you fear portability issues, you can use the % operator.

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Thanks for the explanation and answers!

Quote:

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Originally Posted by SuperKoko

On the other hand, integer overflow has undefined behavior.

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i thought standards say it will overwrite at 0..
so -5u + 6 will become 1
or is it ub for bitfields.

Quote:

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Originally Posted by SuperKoko

And, if you fear portability issues, you can use the % operator.

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hey i already suggested that :(