Rule 7.0.5, example non-compliant with 7.0.6?

[cgpzs]

Rule 7.0.5 presents this example:

Code:

u8a += static_cast< uint32_t >( u8b ) // Compliant - u8a -> unsigned int

Yes, compliant with 7.0.5. But this code then violates 7.0.6, since there is an implicit narrowing conversion from uint32_t to uint8_t in the assignment, right?

Code:

u8a = (uint32_t)(u8a) + (uint32_t)(u8b);

So, how are we supposed to write this type of code? Like this?

Code:

u8a = static_cast<std::uint8_t>(static_cast<std::uint32_t>(u8a) + static_cast<std::uint32_t>(u8b));

Does that really make the code safer?

[misra cpp]

"... Yes, compliant with 7.0.5. But this code then violates 7.0.6, since there is an implicit narrowing conversion from uint32_t to uint8_t in the assignment, right?"

No. The introduction to the section that includes 7.0.5. and 7.0.6 says "Note: compound assignments are not assignments", so 7.0.6 does not apply to a compound assignment.

We recognise that it may be surprising that a += b; is treated differently to a = a + b; but we failed to find an acceptable form of words to say 'treat a = a + ....' as 'a += ....'

The example: u8a += static_cast< uint32_t >( u8b ); is correct and compliant

[cgpzs]

"... Yes, compliant with 7.0.5. But this code then violates 7.0.6, since there is an implicit narrowing conversion from uint32_t to uint8_t in the assignment, right?"

No.  The introduction to the section that includes 7.0.5. and 7.0.6 says "Note: compound assignments are not assignments", so 7.0.6 does not apply to a compound assignment.

We recognise that it may be surprising that  a += b;  is treated differently to  a = a + b;  but we failed to find an acceptable form of words to say  'treat a = a + ....'  as  'a += ....'

The example:  u8a += static_cast< uint32_t >( u8b );  is correct and compliant

Thanks! I (and my static analyzer supplier  ) totally missed that little bit. As a suggestion, I think it would be good to remind about that directly inside Rule 7.0.6, in a future revision.

[misra cpp]

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