print 8 bits of a byte and &0x80 how it work

Projects Programming
1

Hello!

I want to print 8 bits of a byte on serial monitor, now is work fine !
mbyte = 5 , print result 00000101 is correct.

void setup() {

 Serial.begin(9600);
 
 int num = 5;
 byte mbyte = num;
  
 for (int i = 0; i < 8; i++)  { 
    bool b = mbyte&0x80;
    Serial.print(b); 
    mbyte = mbyte << 1;
}

    Serial.println(); 
}

void loop(){}

but one thing is still not clear !

if i change (mbyte = 5)

bool b = mbyte&0x80;

to

bool b = mbyte;

print result = 11111111
0x80 = 10000000

So i don't understand, why increase &0x80, will get print correct result 00000101 ?

use "bitwise AND" think compare numbers(get a 1 if both bits are 1) for 11111111
i can't get result 00000101.

Thanks! :slight_smile:

Wing

2

hint: tryint b = mbyte;and see what is printed

3

Wingx2:
int num = 5;
byte mbyte = num;[/code]

In programming, I like to maintain variable's data type and its value consistent even by casting if necessary.

1. Your quoted codes are re-written as:

int num = 5;
int mbyte = num;

2. Adding 0x1234 and 0x56

int x1 = 0x1234;
byte x2 = 0x56;

int sum = x1 + (int)x2; 
Serial.print(sum, HEX);
4

I commented on this in another Thread as follows

0x80 is 0b10000000

when you use bitwise AND you get a 1 if both bits are 1, otherwise you get a 0

Using & 0x80 will ensure that all bits except the leftmost bit will be 0. The value of the leftmost bit will depend on the value at that bit position in the other number

...R

5 
bool b = mbyte&0x80;

Is doing more than you expect. It’s equivalent to:

  If ((mybyte&0x80) != 0) b=1;  else b=0

Without the &0x80 to isolate a single bit, 5 will stay non-zero for a long time after shifting...

6

aarg:
hint: try

int b = mbyte;

and see what is printed

Hello:

int b = mbyte;

b=5

with forloop "mbyte = mbyte << 1;"

b=5,10,20,40,80,160.......

GolamMostafa:
In programming, I like to maintain variable's data type and its value consistent even by casting if necessary.

1. Your quoted codes are re-written as:

int num = 5;

int mbyte = num;




**2.** Adding 0x1234 and 0x56

int x1 = 0x1234;
byte x2 = 0x56;

int sum = x1 + (int)x2;
Serial.print(sum, HEX);

Thank you for your experience :slight_smile:

Robin2:
I commented on this in another Thread as follows

...R

Hello!
Yes, I saw it, Thanks :slight_smile:
but Thread is old, i want make new Thread to make clear concept "&0x80".

7

westfw:

bool b = mbyte&0x80;

Is doing more than you expect. It’s equivalent to:

  If ((mybyte&0x80) != 0) b=1;  else b=0

Without the &0x80 to isolate a single bit, 5 will stay non-zero for a long time after shifting...

Thank you for your reply, that makes me clearer :smiley:
I found the information based on your reply

  1. Why if(byte & 0x80) means "If the highest bit in the byte isn't 0?" <<
8

westfw:

bool b = mbyte&0x80;

Is doing more than you expect. It’s equivalent to:

  If ((mybyte&0x80) != 0) b=1;  else b=0

Without the &0x80 to isolate a single bit, 5 will stay non-zero for a long time after shifting...

Hello! One more question

"&0x80"

That "&" mean Bitwise Operators And ??

Thanks!

9

Wingx2:
Hello! One more question

"&0x80"

That "&" mean Bitwise Operators And ??

Thanks!