Complement operator turning uint8_t to signed number

Projects Programming
1

Hi I am Siddhesh and I am an intermediate level arduino user,
I am trying to read PORT C and check for different combinations at port C from the eight relay

My doubt is

when I use complement(~ operator) after left shifting(<< operator) a uint8_t number is just gives me answer in 16 bit form

My Code:-

enum relay_pins {START, UP, DOWN, LEFT, RIGHT, FWD, REV, ALARM};
int port_c=254;
uint8_t spam=1;

void setup()
{ Serial.begin(115200);
   

Serial.println(sizeof(spam));
Serial.println(sizeof(~(spam)));
Serial.println(sizeof(spam<<LEFT));

}

void loop()
{
  
}

Output:-
1
2
2

for my doubt, sorry if this is a silly question but just started to learn in deep of bit manipulation

2

Welcome to the forum

Please follow the advice given in the link below when posting code, in particular the section entitled 'Posting code and common code problems'

Use [color = red]code tags[/color] (the </> icon above the compose window) to make it easier to read and copy for examination

3

What is the question?

4

Why a shift or complement operator converting a unsigned 8 bit integer to signed integer of 16 bits

My Code:-

enum relay_pins {START, UP, DOWN, LEFT, RIGHT, FWD, REV, ALARM};
int port_c=254;
uint8_t spam=1;

void setup()
{ Serial.begin(115200);
   

Serial.println(sizeof(spam));
Serial.println(sizeof(~(spam)));
Serial.println(sizeof(spam<<LEFT));

}

void loop()
{
  
}

Output:-
1
2
2

5

Do not post images of code or Serial monitor output. Follow the advice in the link I posted

6

Done sir, can you please guide me on the issue?

7

This is because of numeric promotion. Bitwise operators work on integers, when smaller types are used they are automatically promoted.

https://en.cppreference.com/w/cpp/language/implicit_conversion#Numeric_promotions

[edit]

By the way, here is a trick that may help when you want to know the type of a value.

template <class T> void show(T) {
  Serial.println(__PRETTY_FUNCTION__);
}

void setup() {
  // ...
  show('a');
  show('a' << 1);
}

This results in:

void show(T) [with T = char]
void show(T) [with T = int]
1 Like
8

Thanks, it really helped me understand, I worked around it by just converting it again into a uint8_t after doing my << and ~ operations.
Is there any more effecient way?

My workaround:-

uint8_t spam=1;

Serial.println(sizeof(spam));  // 8bit unsigned

Serial.println(sizeof(~(spam)));  // 16 bit signed

Serial.println(sizeof(spam<<LEFT));  // 16 bit signed

Serial.println (uint8_t(~(spam<<START))) ; // 8 bit unsigned
9

not too sure what you're are trying to achieve, but simply writing this as code ensures that 'spam' is always an 8bit unsigned value...

enum relay_pins {START, UP, DOWN, LEFT, RIGHT, FWD, REV, ALARM};

void setup() {
  // put your setup code here, to run once:
  Serial.begin(115200);

uint8_t spam=1;

Serial.println(sizeof(spam));  // 8bit unsigned

spam = ~spam;

Serial.println(sizeof(spam)); //8 bit unsigned

spam<<=LEFT;

Serial.println(sizeof(spam));  //8 bit unsigned

spam = ~(spam<<START);

Serial.println (sizeof(spam)); //8 bit unsigned
}

void loop() {
  // put your main code here, to run repeatedly:

}

hope that helps....

1 Like
10

Not that I know of. It will be converted back to an uint8_t as soon as you assign it to a variable of that type.

If it really bothers you, perhaps you can use a function, e.g.,

uint8_t shl(uint8_t const v, size_t const n) {
  return v << n;
}
1 Like
11

Works great!!

12

Thanks for the help got it :slight_smile:

13

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