If I had a integer n, what does the statement n>>=1 do? convert n to binary, then shift to right one bit? or n stays in decimal format then gets shifted? Thanks.
If I had a integer n, what does the statement n>>=1 do? convert n to binary, then shift to right one bit? or n stays in decimal format then gets shifted? Thanks.
Convert to binary? Numbers are always stored like that. There is nu such thing as converting a variable to decimal or hex, they are always stored as binary numbers. The numerical base does only matter when you convert the int to a string.
To answer your question: n>>=1 shifts the bits of n to the right one bit.
Binary (base 2) is just a way of representing data, as decimal (base 10) and hexadecimal (base 16) are.
So n can be represented at a bit-level in binary, everything can.
n >> 1 Will shift every bit in n one position right...
for example, if n = "0100 0101".
then after n >> 1, n = "0010 0010"
As you notice, 0 is used as padding...
You may like to read: http://www.cprogramming.com/tutorial...operators.html
Being picky...
For any integer n and b, when n >>= b is executed, n will be right shifted b bits and assigned the result. You will not notice a change in n's radix. All integers have a binary format in memory anyway.
Sorry to interject, is that the C equivalent for the ASM command SHR?
Yes, except, the result isn't assigned.
This leads be to a related question,
how does cin and cout work? is << and >> not bitshift left/right? Just curious...
I've searched for such an explanation with no results.
Depends upon if the n is signed or not.
If n is signed:Code:n>>=1
If n is unsigned:Code:mov eax, n sar eax, 1 mov n, eax
Code:mov eax, n shr eax, 1 mov n, eax<< and >> are just operators. In C++, you can overload operators. Since it wouldn't make as much sense to bit shift streams, they've been overloaded to accept input and output, respectively.Originally Posted by zacs7
Ah, me see. Thanks
