how do i write this

[M.Manjunatha]

How do i write this in a simple fashion?? (a & b) ! = b

 

[joecook]

How do i write this in a simple fashion?? (a & b) ! = b

That looks pretty sinple to me.

Do you like: (a ^ b) & a

better?

Joe Cook

 

[joecook]

How do i write this in a simple fashion?? (a & b) ! = b

That looks pretty sinple to me.

Do you like: (a ^ b) & b

better?

Joe Cook

 

[Pascal J. Bourguignon]

How do i write this in a simple fashion?? (a & b) ! = b

This is not legal C or C++ syntax.

The ! operator is not a post-operator but a prefix operator.
An expression is not a valid l-value for the = operator.

But supposing you meant (a&b)!=b, how simplier do you imagine it could become?
You can always introduce function abstractions (and I'd advise you to do so),
but is it really simplier?

This expression test whether b has set any bit not set in a.

That is, considering a and b to represent sets of small integers,
whether b is a strict superset of a,
ie. whether b is not a subset of a.

So you could abstract the expression away as:

bool is_strict_superset_of(int super,int set){ return((set & super) != super); }
// or:
bool is_not_subset_of(int sub,int set){ return((set & sub) != sub); }


Is is_strict_superset_of(b,a) simplier than (a&b)!=b ?
I think so, but do you agree?

 

[drugz']

How do i write this in a simple fashion?? (a & b) ! = b

Simple: (!a)

 

[alasham.said]

( a & b ) != b
=> [( a & b ) & !b ] | [!( a & b ) & b ]
=> [ ( a & !b ) & ( b & !b ) ] | [( !a | !b ) & b ]
=> [ ( a & !b ) & F ] | [ ( !a & b ) | ( !b & b ) ]
=> [ F ] | [ ( !a & b ) | F ]
=> [ ( !a & b ) | F ]
=> !a & b

But this is more