ROTL and ROTR for 64 bits

Hi, is there a Circular left bit shift operator for 64 bits? for exemple I have this number x=9223372036855037954 circular left shift and right shift by 4 bits should give the following results: xr=2882303761517133824 xl=4194344 thanks a lot

Answer 1

[Jan]

a = uint64(9223372036855037954);  % Not correct in 2009a!
n = rem(n, 64);
bitor(bitshift(a, -n), bitshift(a, (64-n)))
bitor(bitshift(a, n),  bitshift(a, (n-64)))

[suzi freunden]

thanks a lot Jan

the result that I get are:

ans =17870283321406144511

ans =4194296

but not the expected one which is :

2882303761517133824 and 4194344

[Jan]

See: http://www.mathworks.com/matlabcentral/answers/9886-uint64-equivalence

In 2009a, "a=uint64(9223372036855037954)" leads to: a = 9223372036855037951. Unfortunately the number is converted to a DOUBLE at first loosing some bits of precision, and cast to UINT64 afterwards. This behavior has been made consistent in 2010b. The UINT64 support in 2009a is very limited, e.g. operators are not implemented.

Please post your Matlab version.

[suzi freunden]

yes I notice this cast, my matlab version is 2009b....

Answer 2

[Walter Roberson]

No there is not. Extract the N bits at the edge using bitget(), use bitshift() to do the shift, and then bitset() the extra bits into place at the other edge.

[suzi freunden]

thanks a lot but bitget have a maximum less than 2^53

[Walter Roberson]

bitget() does not have such a restriction in R2009b.

http://www.mathworks.com/help/releases/R2009b/techdoc/ref/bitget.html

However, you may have difficulty constructing an unsigned numeric object with higher-order bits set in R2009b. That is a problem with the MATLAB parser, not with the operation of bitget.