From: already5chosen@yahoo.com   
      
   On Mon, 24 Nov 2025 05:06:33 -0800   
   Keith Thompson  wrote:   
      
   > David Brown  writes:   
   > [...]   
   > > Yes, exactly.  At the call site, the size of the _BitInt type is   
   > > always a known compile-time constant, so it can easily be passed on.   
   > > Thus :   
   > >   
   > > 	_BitInt(N) x;   
   > > 	_BitInt(M) y;   
   > > 	_BitInt(NM) z = x * y;   
   > >   
   > > can be implemented as something like :   
   > >   
   > > 	__bit_int_signed_mult(NM, (unsigned char *) &z,   
   > > 			N, (const unsigned char *) &x,   
   > > 			M, (const unsigned char *) &y);   
   >   
   > That looks like it's supposed to avoid overflow (I'm assuming NM is N   
   > + M), but it wouldn't work.  The type of a C expression is almost   
   > always determined by the expression itself, regardless of the context   
   > in which it appears. The type of x * y is _BitInt(max(N, M)), not   
   > _BitInt(N+M), so it can overflow even if the full result would fit   
   > into z.   
   >   
   > You can do this instead (not tested):   
   >   
   >     _BitInt(N) x;   
   >     _BitInt(M) y;   
   >     _Bit_Int(N+M) z = (_BitInt(N+M))x * y;   
   >   
   > (I'm assuming N+M is sufficient, but I might have missed an off-by-one   
   > error somewhere.)   
   >   
      
   You missed nothing. N+M is both sufficient and necessary. The latter   
   because of -(2**(N-1)) * -(2**(M-1)).   
      
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