Rule 6-2-1 Assignment operators shall not be used in subexpressions

[nishiyama]

Hello.

Why are the following MISRA Example compliant?
Is not a sub-expression?

Code:

if ( int16_t i = foo ( ) ) // Compliant
{
}

It looks like the following, but the following are non-compliant.
What are the differences?

Code:

if ( x = y ) // Non-compliant
{
　foo ( );
}

[dg1980]

IMHO, both are technically sub-expressions.
It's either a mistake that the second one is compliant or the intent of MISRA was to add an explicit exception for declaration+initialization of block scope variables like i (x is not, so you have the side effect of changing its value + confusing it with ==).
Anyway, an official clarification would be nice.

[misra cpp]

Firstly, we agree with nishiyama’s interpretation of 6-2-1, the first example is compliant and the second isn’t. The technical reason for why the first is compliant is that it doesn’t contain an assignment (as defined by the C++ standard). When an object is declared, what appears to be an assignment is an initialisation. The C++ standard defines different behaviours for assignment and initialisation (e.g. you cannot assign to a const object, but you can – indeed must – initialise it), so MISRA C++ makes the same distinction.

The more practical reason is that there is no reason to ban if ( int16_t i = foo ( ) ) as there is no possibility of unexpected or undefined behaviour and it would be difficult to ban without also banning for ( int16_t i = 0; … which is such a common coding idiom that it has to be allowed.