Bit shifting two bytes into one signed int

Using Arduino Programming Questions
1

I am reading audio wav file data from SD.

Two bytes represent one signed 16 bit Int. (includes negative numbers).
My code works for positive numbers but not for negatives.

byte a = 16; byte b =0;
int ret = (b << 8) + a ;
ret becomes 16. this is CORRECT. Horray!

byte a = 253; byte b = 255;
int ret = (b << 8) + a ;
ret becomes 65533. this is NOT correct.... ret should be -3. Arrrrrg.

I read about the "Two's complement" method of signed binary representation of negatives but can't seem to wrap my head around it.

2

Show how you print the int variable.

4

int intSigned() {
byte a = f.read();
byte b = f.read();

  Serial.print(a); Serial.print(","); Serial.print(b);
  int r = (b << 8) + a ;

  Serial.print("="); Serial.print(r); Serial.print("..."); 
  return r;

}

I also have a program in C# on my PC that is giving me the correct numbers for the same data using .net's BitConverter.ToInt16() function. So i know the indexes i am reading are correct.

5

You mean char ± 128/127 ?

6

Which platform are you using? Is the int data type 32 or 16 bit ?

7

standard 16 bit.. signed with the industry standard "two's complement" method. I believe its universal across platforms.

8

Problem is: an int can have different sizes on different platforms. If you want truly universal datatypes, use something like int16_t: that is guaranteed to have exactly 16 bits, everywhere.

Take a look at this (tested on a PC):

#include <stdio.h>
#include <stdint.h>

uint8_t a = 253, b = 255;
int ret;
int16_t ret_16;

int main (void)
{
    ret_16 = ret = (b << 8) + a;
    printf ("An int is %lu bytes long.\n", sizeof (int));
    printf ("An int16_t is %lu bytes long.\n",sizeof (int16_t));
    printf ("The value as int is: %d\n", ret);
    printf ("The value as int16_t is: %d\n", ret_16);
    
    ret = (int16_t) ((b << 8) + a);
    printf ("With typecast, the int value is: %d\n", ret);
    
    return 0;
}

Output:

An int is 4 bytes long.
An int16_t is 2 bytes long.
The value as int is: 65533
The value as int16_t is: -3
With typecast, the int value is: -3

So, either save the value into an int16_t or use typecast.

9

Arduinos are not “standard” either. Which Arduino did you run your test program on?

10

What do you think the compiler will do with this?

Will it shift the byte b 8 bits to the left, effectively making b zero before adding a and converting the result to an int, or will it convert b to an int first before carrying out the shift?

11

65533 and -3 are the same bit pattern, if you only have 16 bits (say, on an AVR)
Since "byte" is unsigned, the expression "(b<<8) + 1" will also be treated as unsigned, and won't be sign-extended when assigned to a 32bit "signed int"

Try:

 x = (int16_t)(b << 8) + a ;

works on a desktop:

#include <stdio.h>
#include <stdint.h>

uint8_t a=253, b=255;

int main() {
  int x = (b << 8) + a ;
  printf("result of int x = (b << 8) + a = %d\n", x);
  x = (int16_t)(b << 8) + a ;
  printf("result of     x = (int16_t)(b << 8) + a = %d\n", x);
}

 ./a.out
result of int x = (b << 8) + a = 65533
result of     x = (int16_t)(b << 8) + a = -3
12

Look into using the union structure for this.

13

Bullet Proof:

void setup() {
  uint16_t temp;
  int16_t ret;
  uint8_t a = 253;
  uint8_t b = 255;

  Serial.begin(115200);
  delay(1000);

  temp = ((uint16_t) b) << 8 | a;
  memcpy (&ret, &temp, sizeof(int16_t));
  Serial.println(ret);
}

void loop() {
}

Serial Output:

-3
14

Thank you so much. Your answer seems to be working for me!
Actually, seem I only needed the one cast. here is the final code i am using

  byte a = f.read();
  byte b = f.read();  
   int r = (int16_t)(b << 8) + a ;

and thanks to everyone else for responding as well.

15

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