32 bit binary to 24 bit signed conversion - c++

Question

I'm using a 24 bit I2C ADC with the Arduino and there is no 3 byte (24 bit) data type so I instead used the uint32_t which is a 32 bit unsigned int. My actual output however, is a 24 bit signed number as you can see below:
Also here is the code that I used to read the results if you're interested:
uint32_t readData(){
Wire.beginTransmission(address);
Wire.write(0x10);
Wire.endTransmission();
Wire.requestFrom(address,3);
byte dataMSB = Wire.read();
byte data = Wire.read();
byte dataLSB = Wire.read();
uint32_t data32 = dataMSB;
data32 <<= 8;
data32 |= data;
data32 <<= 8;
data32 |= dataLSB;
return data32;
}
In order for this number to be useful, I need to convert it back to a 24 bit signed integer (I'm not sure how to do that or eve if it's possible because 24 is not a power of 2) so I'm a bit stuck. It would be great if somebody can help me as I'm almost finished with the project and this is one of the last few steps.

Answer 1

The problem is that there’s no safe and portable way to use shifting for sign extension in C — at best it is implementation defined. So if you want to do it portably, you need to convert your 2s-complement value manually into a signed integer.
int32_t cvt24bit(uint32_t val) {
val &= 0xffffff; // limit to 24 bits -- may not be necessary
if (val >= (UINT32_C(1) << 23))
return (int32_t)val - (INT32_C(1) << 24);
else
return val;
}
this will take your 24-bit two’s-complement value in a uint32_t and convert it to a (signed) int32_t.

Answer 2

Conversion from 24-bit two’s complement in a uint32_t to int32_t can be done with:
int32_t Convert(uint32_t x)
{
int32_t t = x & 0xffffff;
return t - (t >> 23 << 24);
}
The x & 0xffffff ensures the number has no spurious bits above bit 23. If it is certain no such bits are set, then the statement can be just int32_t t = x;.
Then t >> 23 removes bits 0 to 22, leave just bit 23, which is the sign bit for a 24-bit integer. Then << 24 scales this, producing either 0 (for positive numbers) or 224 (for negative numbers). Subtracting that from t produces the desired value.

Answer 3

Use int32_t instead of unit32_t for data32. Then before returning the value, shift it left by 8, then right by 8 to sign extend it.
So this code:
uint32_t readData(){
Wire.beginTransmission(address);
Wire.write(0x10);
Wire.endTransmission();
Wire.requestFrom(address,3);
byte dataMSB = Wire.read();
byte data = Wire.read();
byte dataLSB = Wire.read();
int32_t data32 = dataMSB;
data32 <<= 8;
data32 |= data;
data32 <<= 8;
data32 |= dataLSB;
return (data32 << 8) >> 8;
}