line of code.

E

electricity86

New Member

• Apr 1, 2009

• #1

What does this line mean:

P0IFG = (HAL_KEY_P0INT_LOW_USED | HAL_KEY_POINT_HIGH_USED);

Where:
#define HAL_KEY_P0INT_LOW_USED 0
#define HAL_KEY_POINT_HIGH_USED 0

What will be inside P0IFG after that line, if before that line, his value was xxxxxxxx (eight 'x's).

Thanks.

 

Papabravo

Well-Known Member

• Apr 1, 2009

• #2

Should be zeros!
Was this a trick question?

 

Last edited: Apr 1, 2009

E

electricity86

New Member

• Apr 1, 2009

• #3

no, thanks.

So if HAL_KEY_POINT_HIGH_USED was equal 1,
Then the value of P0IFG
would be 0x0F?

 

Sceadwian

Banned

• Apr 1, 2009

• #4

0 OR 0 = 0
I'm not sure why you're having a problem with this question?

 

DirtyLude

Well-Known Member

• Apr 1, 2009

• #5

I guess,

1. there's no such numeric value as 8 x's unless it was something like base 40 or something.

2. The pipe character '|' is a bitwise or operator in C. Google 'bitwise or' and you'll get an explanation of how it works.

 

Sceadwian

Banned

• Apr 1, 2009

• #6

In digital logic x's are commonly used to indicated an 'unknown' state.

 

Papabravo

Well-Known Member

• Apr 1, 2009

• #7

electricity86 said:

no, thanks.

So if HAL_KEY_POINT_HIGH_USED was equal 1,
Then the value of P0IFG
would be 0x0F?

Click to expand...

No it would be 1
as in 0x01 | 0x00 = 0x01
all day every day