Bit-shift behavior

Question

uint16_t a = 0x00 << 8 + 0xB9;
printf("%d",a);

I'm expecting 185 as an output but I'm getting 0.

What is happening here?

https://en.cppreference.com/w/cpp/language/operator_precedence — tkausl, Dec 27 '19 at 12:00
@tkausl Was just about to post the exact same link! Note: `+` has higher precedence than `<<` - so you are shifting `0x00` by `n` bits. — Adrian Mole, Dec 27 '19 at 12:01

score 5 · Accepted Answer · answered Dec 27 '19 at 12:01

5

If you look at this link, you'll see that the order of precedence means that the addition is performed before the shift. Change your code to

uint16_t a = (0x00 << 8) + 0xB9;
printf("%d",a);

to see the desired behaviour.

answered Dec 27 '19 at 12:01

1 Answers1