Will bit-shift by zero bits work correctly?

Question

Say I have a function like this:

inline int shift( int what, int bitCount )
{
    return what >> bitCount;
}

It will be called from diffe

Answer 1

To make the function somewhat self documenting, you may want to change bitCount to unsigned to signify to callers that a negative value is not valid.

Answer 2

The compiler could only perform this optimisation do that if it knew at compile time that the bitCount value was zero. That would mean that the passed parameter would have to be a constant:

const int N = 0;
int x = shift( 123, N );

C++ certainly allows such an optimisation to be performed, but I'm not aware of any compilers that do so. The alternative approach the compiler could take:

int x = n == 0 ? 123 : shift( 123, n );

would be a pessimisation in the majority of cases and I can't imagine compiler writer implementing such a thing.

Edit: AA shift of zero bits is guaranteed to have no effect on the thing being shifted.

Answer 3

It will work correctly on any widely used architecture (I can vouch for x86, PPC, ARM). The compiler will not be able to reduce it to a noop unless the function is inlined.

Answer 4

About the correctness of arg << 0 or arg >> 0, no problem, absolutely fine.

About the eventual optimizations: This will not be reduced to a >nop< when called with a constant what=0 and/or bitcount=0, unless you declare it as inline and choose optimizations (and your compiler of choice understands what inline is).

So, bottom line, optimize this code by conditionally calling the function only if the OR of arguments is non zero (about the fastest way I figure to test that both args are non-zero).

Answer 5

It is certain that at least one C++ compiler will recognize the situation (when the 0 is known at compile time) and make it a no-op:

Source

inline int shift( int what, int bitcount)
{
  return what >> bitcount ;
}

int f() {
  return shift(42,0);
}

Compiler switches

icpc -S -O3 -mssse3 -fp-model fast=2 bitsh.C

Intel C++ 11.0 assembly

# -- Begin  _Z1fv
# mark_begin;
       .align    16,0x90
        .globl _Z1fv
_Z1fv:
..B1.1:                         # Preds ..B1.0
        movl      $42, %eax                                     #7.10
        ret                                                     #7.10
        .align    16,0x90
                                # LOE
# mark_end;
        .type   _Z1fv,@function
        .size   _Z1fv,.-_Z1fv
        .data
# -- End  _Z1fv
        .data
        .section .note.GNU-stack, ""
# End

As you can see at ..B1.1, Intel compiles "return shift(42,0)" to "return 42."

Intel 11 also culls the shift for these two variations:

int g() {
  int a = 5;
  int b = 5;
  return shift(42,a-b);
}

int h(int k) {
  return shift(42,k*0);
}

For the case when the shift value is unknowable at compile time ...

int egad(int m, int n) {
  return shift(42,m-n);
}

... the shift cannot be avoided ...

# -- Begin  _Z4egadii
# mark_begin;
       .align    16,0x90
        .globl _Z4egadii
_Z4egadii:
# parameter 1: 4 + %esp
# parameter 2: 8 + %esp
..B1.1:                         # Preds ..B1.0
        movl      4(%esp), %ecx                                 #20.5
        subl      8(%esp), %ecx                                 #21.21
        movl      $42, %eax                                     #21.10
        shrl      %cl, %eax                                     #21.10
        ret                                                     #21.10
        .align    16,0x90
                                # LOE
# mark_end;

... but at least it's inlined so there's no call overhead.

Bonus assembly: volatile is expensive. The source ...

int g() {
  int a = 5;
  volatile int b = 5;
  return shift(42,a-b);
}

... instead of a no-op, compiles to ...

..B3.1:                         # Preds ..B3.0
        pushl     %esi                                          #10.9
        movl      $5, (%esp)                                    #12.18
        movl      (%esp), %ecx                                  #13.21
        negl      %ecx                                          #13.21
        addl      $5, %ecx                                      #13.21
        movl      $42, %eax                                     #13.10
        shrl      %cl, %eax                                     #13.10
        popl      %ecx                                          #13.10
        ret                                                     #13.10
        .align    16,0x90
                                # LOE
# mark_end;

... so if you're working on a machine where values you push on the stack might not be the same when you pop them, well, this missed optimization is likely the least of your troubles.

Answer 6

According to K&R "The result is undefined if the right operand is negative, or greater than or equal to the number of bits in the left expression's type." (A.7.8) Therefore >> 0 is the identity right shift and perfectly legal.