Can someone explain this?

[Pommie]

I know there's got to be a simple explanation but I can't see it,
I thought all these would be the same, I'm trying to put 51 in the highest byte of work.

uint32_t work;

  work=51*(2^24);

  work=51<<24;

  work=51;
  work<<=24;

but when the result is printed out, I get,
1326
0
855638016
The actual code,

uint32_t work;

void setup(){
  Serial.begin(115200);
  work=51*(2^24);
  Serial.println(work);
  work=51<<24;
  Serial.println(work);
  work=51;
  work<<=24;
  Serial.println(work);
}

void loop(){
  
}

What am I missing?

Thanks,

Mike.

 

[rjenkinsgb]

Try using 51L to force a long constant?

Any intermediate math may be done as integer, otherwise.

 

[Pommie]

Yup, that fixed it except for the first one, even doing,
work=51L*(2L^24L);
And all other combos,
Doesn't fix it - answer is 1326!!!!

Mike.

 

[rjenkinsgb]

What do you get for
work=(2L^24L);

 

[Pommie]

What do you get for
work=(2L^24L);

Now, this is weird, I get 26.

Mike.

 

[Pommie]

And also for,
work=(2L^24);
and
work=(2^24);
It appears that the compiler is taking ^ as +.
Just tried,
work=(2^20);
and it gives 22!!!


Mike.

 

[rjenkinsgb]

Looks like a compiler bug!
It explains the 1326 result, at least.

Does the same thing happen with small powers, eg. 2^3 ?

 

[Ian Rogers]

^ is a bitwise xor on the arduino ...

(51<<24) is the way they do it or pow();

 

[Pommie]

^ is a bitwise xor on the arduino ...

Knew I was missing something.

(51<<24) is the way they do it or pow();

Actually, that gives zero, As pointed out above, it should be,
(51L<<24)

Just kinda unlucky that 20 xor 2 is 22.

Anyway, problem solved.

Thanks all,

Mike.
Edit, note to self, stop using Excel terminology.

 

[Ian Rogers]

Yup! just checked.. constants are 16 bit unless u, l or ul are specified