Unexpected bit shifting result

Question

I have:

(gdb) display/t raw_data[4]<<8
24: /t raw_data[4]<<8 = 1111100000000
(gdb) display/t raw_data[5]
25: /t raw_data[5] = 11100111
(gdb) display/t (raw_data[4]<<8)|raw_data[5]
26: /t (raw_data[4]<<8)|raw_data[5] = 11111111111111111111111111100111

Why is the result on line 26 not 0001111111100111 ? Thanks.

edit: More specifically:

(gdb) display/t raw_data[5]
27: /t raw_data[5] = 11100111
(gdb) display/t 0|raw_data[5]
28: /t 0|raw_data[5] = 11111111111111111111111111100111

Why is the result on line 26 not 11100111 ?

Answer 1

Your data type is a char , which on your platform appears to be signed. The entry raw_data[5] holds the negative number -25.

The print format t prints the data as unsigned integer in binary. When you print raw_data[5] , it is converted to the unsigned char 213, but has only 8 bits. When you do the integer arithmetic on the data, the chars are promoted to a 32-bit integer.

Promoting the negative char value -25 to a signed int will, of course, yield -25, but its representation as an unsigned int is now 2^^32 + x , whereas as an unsigned char it was 2^^8 + x . That's where all the ones at the beginning of the 32-bit binary number come from.

It's maybe better to work with unsigned raw data.

Answer 2

Let's just ignore the first block, since the second block is a minimal reproduction.

Also note that 0 | x 0 | x preserves the value of x , but causes the usual integral promotions.

Then the second block is not so unexpected.

(gdb) display/t raw_data[5]
27: /t raw_data[5] = 11100111

Ok, raw_data[5] is int8_t(-25)

(gdb) display/t 0|raw_data[5]
28: /t 0|raw_data[5] = 11111111111111111111111111100111

and 0|raw_data[5] is int(-25) . Indeed, the value was preserved.

Answer 3

The constant 8 caused a promotion to a signed integer, so you're seeing sign extension as well at the promotion. Change it to UINT8_C(8). You'll need to include stdint.h for the macro.