C++ How to combine two signed 8 Bit numbers to a 16 Bit short? Unexplainable results

c++ byte signed short
Natalie picture Natalie · Nov 24, 2011 · Viewed 50.7k times · Source

I need to combine two signed 8 Bit _int8 values to a signed short (16 Bit) value. It is important that the sign is not lost.

My code is:

 unsigned short lsb = -13;
 unsigned short msb = 1;
 short combined = (msb << 8 )| lsb;

The result I get is -13. However, I expect it to be 499.

For the following examples, I get the correct results with the same code:

msb = -1; lsb = -6; combined = -6;
msb = 1; lsb = 89; combined = 345; 
msb = -1; lsb = 13; combined = -243;

However, msb = 1; lsb = -84; combined = -84; where I would expect 428.

It seems that if the lsb is negative and the msb is positive, something goes wrong! What is wrong with my code? How does the computer get to these unexpected results (Win7, 64 Bit and VS2008 C++)?

Answer

Retired Ninja picture Retired Ninja · Nov 24, 2011

Your lsb in this case contains 0xfff3. When you OR it with 1 << 8 nothing changes because there is already a 1 in that bit position.

Try short combined = (msb << 8 ) | (lsb & 0xff);

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