Question about bitwise and logical and operator

[Arnold]

Hi Charles,

I do not know much about logical and bitwise operators, so I am not sure how to interpret my results. I tried this little code:

Code: [Select]

int r1,r2,r3,r4,r5,r6
int a=60, b=13
 
r1 = (a and b)    'bitwise and
r2 = (a & b)      'bitwise and

r3 = (60 && 13)   'Logical and

r4 = (a&b)        'addition?

r5 = (a& b)    'bitwise and
r6 = (a &b)    'address?

print r1 ", " r2 ", " r3 ", " r4 ", " r5 ", " r6

r4 seems to be an addition and r6 seems to point to an address. Are these the expected results?

Roland

[Charles Pegge]

Hi Roland,

Very interesting.

What the compiler sees in the case of r4:

r4 = (a&  b)

the '&' is seen as a suffix of a, like a$, which is stripped off.

r4 = (a  b)

Since '+' is the default operator, this is interpreted as:

r4 = (a+b)

&b is interpreted as an address in r6

The & symbol does too many things

[Arnold]

Thank you Charles. I would normaly use (a and b), yet in this case I copied a formula coded in C and forgot to do this. But now I understand what happened.

[Charles Pegge]

Perhaps I should alter the behaviour of r6.