Are the shift operators arithmetic or logical in C

Question

In C  are the shift operators   lt  lt    gt  gt   arithmetic or logical

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When shifting left  there is no difference between arithmetic and logical shift  When shifting right  the type of shift depends on the type of the value being shifted    As background for those readers unfamiliar with the difference  a  logical  right shift by 1 bit shifts all the bits to the right and fills in the leftmost bit with a 0  An  arithmetic  shift leaves the original value in the leftmost bit  The difference becomes important when dealing with negative numbers    When shifting an unsigned value  the    operator in C is a logical shift  When shifting a signed value  the    operator is an arithmetic shift   For example  assuming a 32 bit machine   signed int x1   5  assert  x1  gt  gt  1     2   signed int x2   -5  assert  x2  gt  gt  1     -3   unsigned int x3    unsigned int -5  assert  x3  gt  gt  1     0x7FFFFFFD

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When you do   - left shift by 1 you multiply by 2  - right shift by 1 you divide by 2   x   5  x  gt  gt  1  x   2   x 5 2    x   5  x  lt  lt  1  x   10  x 5 2

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Well  I looked it up on wikipedia  and they have this to say      C  however  has only one right shift   operator      Many C compilers choose   which right shift to perform depending   on what type of integer is being   shifted  often signed integers are   shifted using the arithmetic shift    and unsigned integers are shifted   using the logical shift    So it sounds like it depends on your compiler   Also in that article  note that left shift is the same for arithmetic and logical   I would recommend doing a simple test with some signed and unsigned numbers on the border case  high bit set of course  and see what the result is on your compiler   I would also recommend avoiding depending on it being one or the other since it seems C has no standard  at least if it is reasonable and possible to avoid such dependence

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Left shift  lt  lt    This is somehow easy and whenever you use the shift operator  it is always a bit-wise operation  so we can t use it with a double and float operation  Whenever we left shift one zero  it is always added to the least significant bit  LSB    But in right shift  gt  gt  we have to follow one additional rule and that rule is called  sign bit copy   Meaning of  sign bit copy  is if the most significant bit  MSB  is set then after a right shift again the MSB will be set if it was reset then it is again reset  means if the previous value was zero then after shifting again  the bit is zero if the previous bit was one then after the shift it is again one  This rule is not applicable for a left shift    The most important example on right shift if you shift any negative number to right shift  then after some shifting the value finally reach to zero and then after this if shift this -1 any number of times the value will remain same  Please check

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Here are functions to guarantee logical right shift and arithmetic right shift of an int in C   int logicalRightShift int x  int n        return  unsigned x  gt  gt  n    int arithmeticRightShift int x  int n        if  x  lt  0  amp  amp  n  gt  0          return x  gt  gt  n      0U  gt  gt  n       else         return x  gt  gt  n

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According to K amp R 2nd edition the results are implementation-dependent for right shifts of signed values   Wikipedia says that C C    usually  implements an arithmetic shift on signed values   Basically you need to either test your compiler or not rely on it  My VS2008 help for the current MS C   compiler says that their compiler does an arithmetic shift

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In terms of the type of shift you get  the important thing is the type of the value that you re shifting  A classic source of bugs is when you shift a literal to  say  mask off bits  For example  if you wanted to drop the left-most bit of an unsigned integer  then you might try this as your mask    0  gt  gt  1   Unfortunately  this will get you into trouble because the mask will have all of its bits set because the value being shifted   0  is signed  thus an arithmetic shift is performed  Instead  you d want to force a logical shift by explicitly declaring the value as unsigned  i e  by doing something like this    0U  gt  gt  1

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TL DR Consider i and n to be the left and right operands respectively of a shift operator  the type of i  after integer promotion  be T   Assuming n to be in  0  sizeof i    CHAR BIT      undefined otherwise     we ve these cases    Direction      Type     Value  i    Result                       ----------   --------   ---------   ------------------------     Right   gt  gt     unsigned        0      -8    i    2n                 Right        signed          0      -8    i    2n                 Right        signed         lt  0      Implementation-defined         Left    lt  lt     unsigned        0       i   2n     T MAX   1        Left         signed          0       i   2n                        Left         signed         lt  0      Undefined                       most compilers implement this as arithmetic shift     undefined if value overflows the result type T  promoted type of i  Shifting First is the difference between logical and arithmetic shifts from a mathematical viewpoint  without worrying about data type size  Logical shifts always fills discarded bits with zeros while arithmetic shift fills it with zeros only for left shift  but for right shift it copies the MSB thereby preserving the sign of the operand  assuming a two s complement encoding for negative values   In other words  logical shift looks at the shifted operand as just a stream of bits and move them  without bothering about the sign of the resulting value  Arithmetic shift looks at it as a  signed  number and preserves the sign as shifts are made  A left arithmetic shift of a number X by n is equivalent to multiplying X by 2n and is thus equivalent to logical left shift  a logical shift would also give the same result since MSB anyway falls off the end and there s nothing to preserve  A right arithmetic shift of a number X by n is equivalent to integer division of X by 2n ONLY if X is non-negative  Integer division is nothing but mathematical division and round towards 0  trunc   For negative numbers  represented by two s complement encoding  shifting right by n bits has the effect of mathematically dividing it by 2n and rounding towards -8  floor   thus right shifting is different for non-negative and negative values   for X   0  X  gt  gt  n   X   2n   trunc X    2n  for X  lt  0  X  gt  gt  n   floor X    2n   where    is mathematical division    is integer division  Let s look at an example   37 10   100101 2 37    2   18 5 37   2   18  rounding 18 5 towards 0    10010 2  result of arithmetic right shift  -37 10   11011011 2  considering a two s complement  8-bit representation  -37    2   -18 5 -37   2   -18  rounding 18 5 towards 0    11101110 2  NOT the result of arithmetic right shift  -37  gt  gt  1   -19  rounding 18 5 towards -8    11101101 2  result of arithmetic right shift   As Guy Steele pointed out  this discrepancy has led to bugs in more than one compiler  Here non-negative  math  can be mapped to unsigned and signed non-negative values  C   both are treated the same and right-shifting them is done by integer division  So logical and arithmetic are equivalent in left-shifting and for non-negative values in right shifting  it s in right shifting of negative values that they differ  Operand and Result Types Standard C99   6 5 7   Each of the operands shall have integer types  The integer promotions are performed on each of the operands  The type of the result is that of the promoted left operand  If the value of the right operand is negative or is greater than or equal to the width of the promoted left operand  the behaviour is undefined   short E1   1  E2   3  int R   E1  lt  lt  E2   In the above snippet  both operands become int  due to integer promotion   if E2 was negative or E2   sizeof int    CHAR BIT then the operation is undefined  This is because shifting more than the available bits is surely going to overflow  Had R been declared as short  the int result of the shift operation would be implicitly converted to short  a narrowing conversion  which may lead to implementation-defined behaviour if the value is not representable in the destination type  Left Shift  The result of E1  lt  lt  E2 is E1 left-shifted E2 bit positions  vacated bits are filled with zeros  If E1 has an unsigned type  the value of the result is E1  2E2  reduced modulo one more than the maximum value representable in the result type  If E1 has a signed type and non-negative value  and E1  2E2 is representable in the result type  then that is the resulting value  otherwise  the behaviour is undefined   As left shifts are the same for both  the vacated bits are simply filled with zeros  It then states that for both unsigned and signed types it s an arithmetic shift  I m interpreting it as arithmetic shift since logical shifts don t bother about the value represented by the bits  it just looks at it as a stream of bits  but the standard talks not in terms of bits  but by defining it in terms of the value obtained by the product of E1 with 2E2  The caveat here is that for signed types the value should be non-negative and the resulting value should be representable in the result type  Otherwise the operation is undefined  The result type would be the type of the E1 after applying integral promotion and not the destination  the variable which is going to hold the result  type  The resulting value is implicitly converted to the destination type  if it is not representable in that type  then the conversion is implementation-defined  C99   6 3 1 3 3   If E1 is a signed type with a negative value then the behaviour of left shifting is undefined  This is an easy route to undefined behaviour which may easily get overlooked  Right Shift  The result of E1  gt  gt  E2 is E1 right-shifted E2 bit positions  If E1 has an unsigned type or if E1 has a signed type and a non-negative value  the value of the result is the integral part of the quotient of E1 2E2  If E1 has a signed type and a negative value  the resulting value is implementation-defined   Right shift for unsigned and signed non-negative values are pretty straight forward  the vacant bits are filled with zeros  For signed negative values the result of right shifting is implementation-defined  That said  most implementations like GCC and Visual C   implement right-shifting as arithmetic shifting by preserving the sign bit  Conclusion Unlike Java  which has a special operator  gt  gt  gt  for logical shifting apart from the usual  gt  gt  and  lt  lt   C and C   have only arithmetic shifting with some areas left undefined and implementation-defined  The reason I deem them as arithmetic is due to the standard wording the operation mathematically rather than treating the shifted operand as a stream of bits  this is perhaps the reason why it leaves those areas un implementation-defined instead of just defining all cases as logical shifts

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According to many c compilers     lt  lt  is an arithmetic left shift or bitwise left shift   gt  gt  is an arithmetic right shiftor bitwise right shift

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gcc will typically use logical shifts on unsigned variables and for left-shifts on signed variables  The arithmetic right shift is the truly important one because it will sign extend the variable   gcc will will use this when applicable  as other compilers are likely to do

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GCC does    for -ve -   Arithmetic Shift For  ve  -  Logical Shift

[c] Are the shift operators (<<, >>) arithmetic or logical in C?

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