[c++] Best practices for circular shift (rotate) operations in C++

Left and right shift operators (<< and >>) are already available in C++. However, I couldn't find out how I could perform circular shift or rotate operations.

How can operations like "Rotate Left" and "Rotate Right" be performed?

Rotating right twice here

Initial --> 1000 0011 0100 0010

should result in:

Final   --> 1010 0000 1101 0000

An example would be helpful.

(editor's note: Many common ways of expressing rotates in C suffer from undefined behaviour if the rotate count is zero, or compile to more than just a single rotate machine instruction. This question's answer should document best practices.)

Definitively:

template<class T>
T ror(T x, unsigned int moves)
{
  return (x >> moves) | (x << sizeof(T)*8 - moves);
}

Source Code x bit number

int x =8;
data =15; //input
unsigned char tmp;
for(int i =0;i<x;i++)
{
printf("Data & 1    %d\n",data&1);
printf("Data Shifted value %d\n",data>>1^(data&1)<<(x-1));
tmp = data>>1|(data&1)<<(x-1);
data = tmp;  
}

Assuming you want to shift right by L bits, and the input x is a number with N bits:

unsigned ror(unsigned x, int L, int N) 
{
    unsigned lsbs = x & ((1 << L) - 1);
    return (x >> L) | (lsbs << (N-L));
}

another suggestion

template<class T>
inline T rotl(T x, unsigned char moves){
    unsigned char temp;
    __asm{
        mov temp, CL
        mov CL, moves
        rol x, CL
        mov CL, temp
    };
    return x;
}

If x is an 8 bit value, you can use this:

x=(x>>1 | x<<7);

#define ROTATE_RIGHT(x) ( (x>>1) | (x&1?0x8000:0) )

Below is a slightly improved version of Dídac Pérez's answer, with both directions implemented, along with a demo of these functions' usages using unsigned char and unsigned long long values. Several notes:

1. The functions are inlined for compiler optimizations
2. I used a cout << +value trick for tersely outputting an unsigned char numerically that I found here: https://stackoverflow.com/a/28414758/1599699
3. I recommend using the explicit <put the type here> syntax for clarity and safety.
4. I used unsigned char for the shiftNum parameter because of what I found in the Additional Details section here:

The result of a shift operation is undefined if additive-expression is negative or if additive-expression is greater than or equal to the number of bits in the (promoted) shift-expression.

Here's the code I'm using:

#include <iostream>

using namespace std;

template <typename T>
inline T rotateAndCarryLeft(T rotateMe, unsigned char shiftNum)
{
    static const unsigned char TBitCount = sizeof(T) * 8U;

    return (rotateMe << shiftNum) | (rotateMe >> (TBitCount - shiftNum));
}

template <typename T>
inline T rotateAndCarryRight(T rotateMe, unsigned char shiftNum)
{
    static const unsigned char TBitCount = sizeof(T) * 8U;

    return (rotateMe >> shiftNum) | (rotateMe << (TBitCount - shiftNum));
}

void main()
{
    //00010100 == (unsigned char)20U
    //00000101 == (unsigned char)5U == rotateAndCarryLeft(20U, 6U)
    //01010000 == (unsigned char)80U == rotateAndCarryRight(20U, 6U)

    cout << "unsigned char " << 20U << " rotated left by 6 bits == " << +rotateAndCarryLeft<unsigned char>(20U, 6U) << "\n";
    cout << "unsigned char " << 20U << " rotated right by 6 bits == " << +rotateAndCarryRight<unsigned char>(20U, 6U) << "\n";

    cout << "\n";


    for (unsigned char shiftNum = 0U; shiftNum <= sizeof(unsigned char) * 8U; ++shiftNum)
    {
        cout << "unsigned char " << 21U << " rotated left by " << +shiftNum << " bit(s) == " << +rotateAndCarryLeft<unsigned char>(21U, shiftNum) << "\n";
    }

    cout << "\n";

    for (unsigned char shiftNum = 0U; shiftNum <= sizeof(unsigned char) * 8U; ++shiftNum)
    {
        cout << "unsigned char " << 21U << " rotated right by " << +shiftNum << " bit(s) == " << +rotateAndCarryRight<unsigned char>(21U, shiftNum) << "\n";
    }


    cout << "\n";

    for (unsigned char shiftNum = 0U; shiftNum <= sizeof(unsigned long long) * 8U; ++shiftNum)
    {
        cout << "unsigned long long " << 3457347ULL << " rotated left by " << +shiftNum << " bit(s) == " << rotateAndCarryLeft<unsigned long long>(3457347ULL, shiftNum) << "\n";
    }

    cout << "\n";

    for (unsigned char shiftNum = 0U; shiftNum <= sizeof(unsigned long long) * 8U; ++shiftNum)
    {
        cout << "unsigned long long " << 3457347ULL << " rotated right by " << +shiftNum << " bit(s) == " << rotateAndCarryRight<unsigned long long>(3457347ULL, shiftNum) << "\n";
    }

    cout << "\n\n";
    system("pause");
}

C++20 std::rotl and std::rotr

It has arrived! http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2019/p0553r4.html and should add it to the <bit> header.

cppreference says that the usage will be like:

#include <bit>
#include <bitset>
#include <cstdint>
#include <iostream>

int main()
{
    std::uint8_t i = 0b00011101;
    std::cout << "i          = " << std::bitset<8>(i) << '\n';
    std::cout << "rotl(i,0)  = " << std::bitset<8>(std::rotl(i,0)) << '\n';
    std::cout << "rotl(i,1)  = " << std::bitset<8>(std::rotl(i,1)) << '\n';
    std::cout << "rotl(i,4)  = " << std::bitset<8>(std::rotl(i,4)) << '\n';
    std::cout << "rotl(i,9)  = " << std::bitset<8>(std::rotl(i,9)) << '\n';
    std::cout << "rotl(i,-1) = " << std::bitset<8>(std::rotl(i,-1)) << '\n';
}

giving output:

i          = 00011101
rotl(i,0)  = 00011101
rotl(i,1)  = 00111010
rotl(i,4)  = 11010001
rotl(i,9)  = 00111010
rotl(i,-1) = 10001110

I'll give it a try when support arrives to GCC, GCC 9.1.0 with g++-9 -std=c++2a still doesn't support it.

The proposal says:

Header:

namespace std {
  // 25.5.5, rotating   
  template<class T>
    [[nodiscard]] constexpr T rotl(T x, int s) noexcept;
  template<class T>
    [[nodiscard]] constexpr T rotr(T x, int s) noexcept;

and:

25.5.5 Rotating [bitops.rot]

In the following descriptions, let N denote std::numeric_limits<T>::digits.

template<class T>
  [[nodiscard]] constexpr T rotl(T x, int s) noexcept;

Constraints: T is an unsigned integer type (3.9.1 [basic.fundamental]).

Let r be s % N.

Returns: If r is 0, x; if r is positive, (x << r) | (x >> (N - r)); if r is negative, rotr(x, -r).

template<class T>
  [[nodiscard]] constexpr T rotr(T x, int s) noexcept;

Constraints: T is an unsigned integer type (3.9.1 [basic.fundamental]). Let r be s % N.

Returns: If r is 0, x; if r is positive, (x >> r) | (x << (N - r)); if r is negative, rotl(x, -r).

A std::popcount was also added to count the number of 1 bits: How to count the number of set bits in a 32-bit integer?

In details you can apply the following logic.

If Bit Pattern is 33602 in Integer

1000 0011 0100 0010

and you need to Roll over with 2 right shifs then: first make a copy of bit pattern and then left shift it: Length - RightShift i.e. length is 16 right shift value is 2 16 - 2 = 14

After 14 times left shifting you get.

1000 0000 0000 0000

Now right shift the value 33602, 2 times as required. You get

0010 0000 1101 0000

Now take an OR between 14 time left shifted value and 2 times right shifted value.

1000 0000 0000 0000
0010 0000 1101 0000
===================
1010 0000 1101 0000
===================

And you get your shifted rollover value. Remember bit wise operations are faster and this don't even required any loop.

Since it's C++, use an inline function:

template <typename INT> 
INT rol(INT val) {
    return (val << 1) | (val >> (sizeof(INT)*CHAR_BIT-1));
}

C++11 variant:

template <typename INT> 
constexpr INT rol(INT val) {
    static_assert(std::is_unsigned<INT>::value,
                  "Rotate Left only makes sense for unsigned types");
    return (val << 1) | (val >> (sizeof(INT)*CHAR_BIT-1));
}

The correct answer is following:

#define BitsCount( val ) ( sizeof( val ) * CHAR_BIT )
#define Shift( val, steps ) ( steps % BitsCount( val ) )
#define ROL( val, steps ) ( ( val << Shift( val, steps ) ) | ( val >> ( BitsCount( val ) - Shift( val, steps ) ) ) )
#define ROR( val, steps ) ( ( val >> Shift( val, steps ) ) | ( val << ( BitsCount( val ) - Shift( val, steps ) ) ) )

How abt something like this, using the standard bitset ...

#include <bitset> 
#include <iostream> 

template <std::size_t N> 
inline void 
rotate(std::bitset<N>& b, unsigned m) 
{ 
   b = b << m | b >> (N-m); 
} 

int main() 
{ 
   std::bitset<8> b(15); 
   std::cout << b << '\n'; 
   rotate(b, 2); 
   std::cout << b << '\n'; 

   return 0;
}

HTH,

--- Substituting RLC in 8051 C for speed --- Rotate left carry
Here is an example using RLC to update a serial 8 bit DAC msb first:
                               (r=DACVAL, P1.4= SDO, P1.5= SCLK)
MOV     A, r
?1:
MOV     B, #8
RLC     A
MOV     P1.4, C
CLR     P1.5
SETB    P1.5
DJNZ    B, ?1

Here is the code in 8051 C at its fastest:
sbit ACC_7  = ACC ^ 7 ; //define this at the top to access bit 7 of ACC
ACC     =   r;
B       =   8;  
do  {
P1_4    =   ACC_7;  // this assembles into mov c, acc.7  mov P1.4, c 
ACC     <<= 1;
P1_5    =   0;
P1_5    =   1;
B       --  ; 
    } while ( B!=0 );
The keil compiler will use DJNZ when a loop is written this way.
I am cheating here by using registers ACC and B in c code.
If you cannot cheat then substitute with:
P1_4    =   ( r & 128 ) ? 1 : 0 ;
r     <<=   1;
This only takes a few extra instructions.
Also, changing B for a local var char n is the same.
Keil does rotate ACC left by ADD A, ACC which is the same as multiply 2.
It only takes one extra opcode i think.
Keeping code entirely in C keeps things simpler sometimes.

Overload a function:

unsigned int rotate_right(unsigned int x)
{
 return (x>>1 | (x&1?0x80000000:0))
}

unsigned short rotate_right(unsigned short x) { /* etc. */ }

Most compilers have intrinsics for that. Visual Studio for example _rotr8, _rotr16