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

Question

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.)

Answer 1

See also an earlier version of this answer on another rotate question with some more details about what asm gcc/clang produce for x86.

The most compiler-friendly way to express a rotate in C and C++ that avoids any Undefined Behaviour seems to be John Regehr's implementation. I've adapted it to rotate by the width of the type (using fixed-width types like uint32_t).

#include <stdint.h>   // for uint32_t
#include <limits.h>   // for CHAR_BIT
// #define NDEBUG
#include <assert.h>

static inline uint32_t rotl32 (uint32_t n, unsigned int c)
{
  const unsigned int mask = (CHAR_BIT*sizeof(n) - 1);  // assumes width is a power of 2.

  // assert ( (c<=mask) &&"rotate by type width or more");
  c &= mask;
  return (n<<c) | (n>>( (-c)&mask ));
}

static inline uint32_t rotr32 (uint32_t n, unsigned int c)
{
  const unsigned int mask = (CHAR_BIT*sizeof(n) - 1);

  // assert ( (c<=mask) &&"rotate by type width or more");
  c &= mask;
  return (n>>c) | (n<<( (-c)&mask ));
}

Works for any unsigned integer type, not just uint32_t, so you could make versions for other sizes.

See also a C++11 template version with lots of safety checks (including a static_assert that the type width is a power of 2), which isn't the case on some 24-bit DSPs or 36-bit mainframes, for example.

I'd recommend only using the template as a back-end for wrappers with names that include the rotate width explicitly. Integer-promotion rules mean that rotl_template(u16 & 0x11UL, 7) would do a 32 or 64-bit rotate, not 16 (depending on the width of unsigned long). Even uint16_t & uint16_t is promoted to signed int by C++'s integer-promotion rules, except on platforms where int is no wider than uint16_t.

---

On x86, this version inlines to a single rol r32, cl (or rol r32, imm8) with compilers that grok it, because the compiler knows that x86 rotate and shift instructions mask the shift-count the same way the C source does.

Compiler support for this UB-avoiding idiom on x86, for uint32_t x and unsigned int n for variable-count shifts:

• clang: recognized for variable-count rotates since clang3.5, multiple shifts+or insns before that.
• gcc: recognized for variable-count rotates since gcc4.9, multiple shifts+or insns before that. gcc5 and later optimize away the branch and mask in the wikipedia version, too, using just a ror or rol instruction for variable counts.
• icc: supported for variable-count rotates since ICC13 or earlier. Constant-count rotates use shld edi,edi,7 which is slower and takes more bytes than rol edi,7 on some CPUs (especially AMD, but also some Intel), when BMI2 isn't available for rorx eax,edi,25 to save a MOV.
• MSVC: x86-64 CL19: Only recognized for constant-count rotates. (The wikipedia idiom is recognized, but the branch and AND aren't optimized away). Use the _rotl / _rotr intrinsics from <intrin.h> on x86 (including x86-64).

gcc for ARM uses an and r1, r1, #31 for variable-count rotates, but still does the actual rotate with a single instruction: ror r0, r0, r1. So gcc doesn't realize that rotate-counts are inherently modular. As the ARM docs say, "ROR with shift length, n, more than 32 is the same as ROR with shift length n-32". I think gcc gets confused here because left/right shifts on ARM saturate the count, so a shift by 32 or more will clear the register. (Unlike x86, where shifts mask the count the same as rotates). It probably decides it needs an AND instruction before recognizing the rotate idiom, because of how non-circular shifts work on that target.

Current x86 compilers still use an extra instruction to mask a variable count for 8 and 16-bit rotates, probably for the same reason they don't avoid the AND on ARM. This is a missed optimization, because performance doesn't depend on the rotate count on any x86-64 CPU. (Masking of counts was introduced with 286 for performance reasons because it handled shifts iteratively, not with constant-latency like modern CPUs.)

BTW, prefer rotate-right for variable-count rotates, to avoid making the compiler do 32-n to implement a left rotate on architectures like ARM and MIPS that only provide a rotate-right. (This optimizes away with compile-time-constant counts.)

Fun fact: ARM doesn't really have dedicated shift/rotate instructions, it's just MOV with the source operand going through the barrel-shifter in ROR mode: mov r0, r0, ror r1. So a rotate can fold into a register-source operand for an EOR instruction or something.

---

Make sure you use unsigned types for n and the return value, or else it won't be a rotate. (gcc for x86 targets does arithmetic right shifts, shifting in copies of the sign-bit rather than zeroes, leading to a problem when you OR the two shifted values together. Right-shifts of negative signed integers is implementation-defined behaviour in C.)

Also, make sure the shift count is an unsigned type, because (-n)&31 with a signed type could be one's complement or sign/magnitude, and not the same as the modular 2^n you get with unsigned or two's complement. (See comments on Regehr's blog post). unsigned int does well on every compiler I've looked at, for every width of x. Some other types actually defeat the idiom-recognition for some compilers, so don't just use the same type as x.

---

Some compilers provide intrinsics for rotates, which is far better than inline-asm if the portable version doesn't generate good code on the compiler you're targeting. There aren't cross-platform intrinsics for any compilers that I know of. These are some of the x86 options:

• Intel documents that <immintrin.h> provides _rotl and _rotl64 intrinsics, and same for right shift. MSVC requires <intrin.h>, while gcc require <x86intrin.h>. An #ifdef takes care of gcc vs. icc, but clang doesn't seem to provide them anywhere, except in MSVC compatibility mode with -fms-extensions -fms-compatibility -fms-compatibility-version=17.00. And the asm it emits for them sucks (extra masking and a CMOV).
• MSVC: _rotr8 and _rotr16.
• gcc and icc (not clang): <x86intrin.h> also provides __rolb/__rorb for 8-bit rotate left/right, __rolw/__rorw (16-bit), __rold/__rord (32-bit), __rolq/__rorq (64-bit, only defined for 64-bit targets). For narrow rotates, the implementation uses __builtin_ia32_rolhi or ...qi, but the 32 and 64-bit rotates are defined using shift/or (with no protection against UB, because the code in ia32intrin.h only has to work on gcc for x86). GNU C appears not to have any cross-platform __builtin_rotate functions the way it does for __builtin_popcount (which expands to whatever's optimal on the target platform, even if it's not a single instruction). Most of the time you get good code from idiom-recognition.

// For real use, probably use a rotate intrinsic for MSVC, or this idiom for other compilers.  This pattern of #ifdefs may be helpful
#if defined(__x86_64__) || defined(__i386__)

#ifdef _MSC_VER
#include <intrin.h>
#else
#include <x86intrin.h>  // Not just <immintrin.h> for compilers other than icc
#endif

uint32_t rotl32_x86_intrinsic(rotwidth_t x, unsigned n) {
  //return __builtin_ia32_rorhi(x, 7);  // 16-bit rotate, GNU C
  return _rotl(x, n);  // gcc, icc, msvc.  Intel-defined.
  //return __rold(x, n);  // gcc, icc.
  // can't find anything for clang
}
#endif

Presumably some non-x86 compilers have intrinsics, too, but let's not expand this community-wiki answer to include them all. (Maybe do that in the existing answer about intrinsics).

---

(The old version of this answer suggested MSVC-specific inline asm (which only works for 32bit x86 code), or http://www.devx.com/tips/Tip/14043 for a C version. The comments are replying to that.)

Inline asm defeats many optimizations, especially MSVC-style because it forces inputs to be stored/reloaded. A carefully-written GNU C inline-asm rotate would allow the count to be an immediate operand for compile-time-constant shift counts, but it still couldn't optimize away entirely if the value to be shifted is also a compile-time constant after inlining. https://gcc.gnu.org/wiki/DontUseInlineAsm.

Answer 2

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));
}

Answer 3

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?

Answer 4

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

Answer 5

Definitively:

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

Answer 6

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

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

Answer 7

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,

Answer 8

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.

Answer 9

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));
}

Answer 10

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 ) ) ) )

Answer 11

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;  
}

Answer 12

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;
}

Answer 13

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");
}

Answer 14

--- 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.

Answer 15

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

Answer 16

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. */ }