Arduino Serial.print（，BIN）奇怪的行为

Question

Code:

char a = 0x70;
char b = 0x80;

Serial.println(a, BIN); // Should print  1110000
Serial.println(b, BIN); // Should print 10000000

Output:

1110000
11111111111111111111111110000000

I know this has something to do with the first bit being one makes it a negative number and maybe it tries to print it as an int by default? Making the char unsigned does not change this, however.

Answer 1

This is inspired by @Ben's comments on the question. It appears that Serial.println((unsigned char)b, BIN); gets the desired output.

Here is my complete sketch:

void setup() {
  Serial.begin(9600);
  // Confirm observations from question
  char a = 0x70;
  char b = 0x80;

  long aPrint = Serial.println(a, BIN); // Should print  1110000
  long bPrint = Serial.println(b, BIN); // Should print 10000000

  // Output println results (Ben comment #1)
  Serial.print("aPrint: ");
  Serial.println(aPrint);
  Serial.print("bPrint: ");
  Serial.println(bPrint);

  // Explicit cast from char
  Serial.print("(int)b: ");
  Serial.println((int)b);

  // Via unsigned char
  Serial.print("(unsigned char)b: ");
  Serial.println((unsigned char)b);
  // And print in binary
  Serial.println((unsigned char)b, BIN);
}

void loop() {
}

Output:

1110000
11111111111111111111111110000000
aPrint: 9
bPrint: 34
(int)b: -128
(unsigned char)b: 128
10000000

Answer 2

char appears to be a signed type on your machine. That means when yout pass it to println , which expects an int , it gets sign-extended to the value you see there. Use a cast as suggested in other answers or switch to unsigned char .

Edit: I noticed your post says that using unsigned doesn't help. That might indicate that the API is doing something funny internally.