CODEWITHSUNDEEP
cmacrosarmbitiar
user1111998
user1111998

Reputation:

Bitwise operators comparison error

compiler: IAR platform: stm32f4

i have this code but it dont work as expected. some one can explain me why?

#define byteGetbit(x,y) ((x) & (0x01 << (y)))

volatile t_uint8 test=0xFF;
if(byteGetbit(test,0x03)==1){ //always false
   printf("hello"); //can't reach the code here
   test = 0;
}

workaround:

if(byteGetbit(test,0x03)){
   printf("hello");
   test = 0;
 }

Upvotes: 0

Views: 437

Answers (3)

Nico Erfurth
Nico Erfurth

Reputation: 3453

You don't read the bit, but you isolate the bit at it's current locations.

You should change byteGetbit to this:

#define byteGetbit(x,y) ((x >> y) & 0x01)

This will shift the bit to bit 0 and then isolate it.

Upvotes: 0

Filipe Gon&#231;alves
Filipe Gon&#231;alves

Reputation: 21223

((x) & (0x01 << (y))) is not a logical operation, thus, it does not yield a 0 or 1 result. In this particular example, ((x) & (0x01 << (y))) results in 0 (for the case that bit y is disabled in x, or it results in a binary number with only bit y set, which can be 1 (if y == 0), but it doesn't have to be 1 (and it never is for the case that y > 0).

If you want byteGetbit to evaluate to either 1 or 0, like logical operations, you should use this instead:

#define byteGetbit(x,y) (((x) & (0x01 << (y))) ? 1 : 0)

This will evaluate to 1 if bit y is set on x and to 0 otherwise. In other words, it gives you the value of the y-th bit on x.

Upvotes: 1

ouah
ouah

Reputation: 145889

byteGetbit yields a value with the corresponding bit set or clear, so to test for the bit set you have to either check:

if(byteGetbit(test,0x03) == 0x03)

or

if(byteGetbit(test,0x03))  // test if byteGetbit(test,0x03) != 0

Upvotes: 2

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