How to get a right shifted number without actual calculation?

Question

I have the following snippet:

int n = 10;
int k = n>>1;
std::cout<<k;

This prints 5.

I want k to be the last digit in binary representation of n. Like bin(n) = 1010 So, I want k to be 0.

I understand long methods are possible. Please suggest a one liner if possible.

Edit:

After going through the comments and answers, I discovered that there are various ways of doing that.

Some of them are:

k = n%2
k = n&1

Thanks to all those who answered the question. :)

“This prints 5.” — it doesn’t print anything. To get the low bit of a value, use `whatever & 0x01`. — Pete Becker, Sep 20 '17 at 15:17
`k = n & 1` or `k = n % 2`. It's basically [getting a single bit](https://stackoverflow.com/q/8695945/995714) — phuclv, Sep 20 '17 at 15:18
`n % 2` produces some trouble with negative `n`, it is a bit odd to say that the least significant bit of a negative number is negative 1 — harold, Sep 20 '17 at 15:23
If you want k to be he last digit in binary representation of n (which is either 0 or 1), you want to know in fact if the number is even or odd ? Is it what you wanted ? if it is the case you can indeed use something line k = n %2 — Mustapha Yousfi, Sep 20 '17 at 15:25
@Actarus> not always. Sometimes you actually want the bit value (from a bit field, from some I/O port, …). Since OP specifically tagged with bit manipulation I guess that's what he wants. — spectras, Sep 20 '17 at 15:44

lakeweb · Answer 1 · 2017-09-20T22:04:03.147

1

int main( )
{
    unsigned int val= 0x1010;
    //so you just want the least siginificant bit?
    //and assign it to another int?
    unsigned int assign= val & 0x1;

    std::cout << assign << std::endl;
    val= 0x1001;
    assign= val & 0x1;
    std::cout << assign << std::endl;
    return 0;
}

UPDATE:

I would add that bit masking is not uncommon with c. I use ints to hold states often

#define STATE_MOTOR_RUNNING     0x0001
#define STATE_UPDATE_DISPLAY    0x0002
#define STATE_COUNTER_READY     0x0004

Then:

unsigned int state= STATE_COUNTER_READY;
if( state & STATE_COUNTER_READY )
{
    start_motor( );
    state|= STATE_MOTOR_RUNNING;
}
//etc...

edited Sep 20 '17 at 22:04

answered Sep 20 '17 at 21:54

lakeweb

1,768
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What's the point of that? I mean, if it was some I/O register, or some serialized form, fine, but for normal code? – spectras Sep 21 '17 at 09:20
HI @spectras, I don't see much of a point, I was just answering the op's question. I added the update to try and frame it in a useful manor. – lakeweb Sep 21 '17 at 15:26

score 0 · Accepted Answer · answered Sep 20 '17 at 15:17

0

You aren't going to be able to avoid some calculation.

int k = n % 10;

will get you the last decimal digit, as that assignment gives k the remainder of division by 10.

answered Sep 20 '17 at 15:17

vacuumhead

449
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OP asked for the last digit in binary though – harold Sep 20 '17 at 15:18
suggest to replace 10 with 2 and add a note that its basically the same as `n & 1` – 463035818_is_not_a_number Sep 20 '17 at 15:19
or even just n% just incase someone wants to work in hex or octal – UKMonkey Sep 20 '17 at 15:30

How to get a right shifted number without actual calculation?

2 Answers2