CODEWITHSUNDEEP
cbit-manipulationbitwise-operators
ivan
ivan

Reputation: 6322

Bit-shift not applying to a variable declaration-assignment one-liner

I'm seeing strange behavior when I try to apply a right bit-shift within a variable declaration/assignment:

unsigned int i = ~0 >> 1;

The result I'm getting is 0xffffffff, as if the >> 1 simply wasn't there. It seems to be something about the ~0, because if I instead do:

unsigned int i = 0xffffffff >> 1;

I get 0x7fffffff as expected. I thought I might be tripping over an operator precedence issue, so tried:

unsigned int i = (~0) >> 1;

but it made no difference. I could just perform the shift in a separate statement, like

unsigned int i = ~0;
i >>= 1;

but I'd like to know what's going on.

update Thanks merlin2011 for pointing me towards an answer. Turns out it was performing an arithmetic shift because it was interpreting ~0 as a signed (negative) value. The simplest fix seems to be:

unsigned int i = ~0u >> 1;

Now I'm wondering why 0xffffffff wasn't also interpreted as a signed value.

Upvotes: 2

Views: 81

Answers (2)

Ian
Ian

Reputation: 30823

It is how c compiler works for signed value. The base literal for number in C is int (in 32-bit machine, it is 32-bit signed int)

You may want to change it to:

unsigned int i = ~(unsigned int)0 >> 1;

The reason is because for the signed value, the compiler would treat the operator >> as an arithmetic shift (or signed shift).

Or, more shortly (pointed out by M.M),

unsigned int i = ~0u >> 1;

Test:

printf("%x", i);

Result:

enter image description here

Upvotes: 1

Danny_ds
Danny_ds

Reputation: 11406

In unsigned int i = ~0;, ~0 is seen as a signed integer (the compiler should warn about that).

Try this instead:

unsigned int i = (unsigned int)~0 >> 1;

Upvotes: 0

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