Logical, arithmetical bitwise shifts
Seeking to clarify something.
It is my understanding that with regard to arithmetical, logical bitwise shifts:
<<
work the same for both>>
shifts differ in that logi开发者_JAVA技巧cal shift will always pad byte with 0, whereas arithmetical shift will pad it with the sign bit.
How can I differentiate this using C?
From what I understand, actual operators are the same <<
,>>
How would command differ between:
int i=1;
printf ("%d\n", i >> 1); // logical shift
int j=1;
printf ("%d\n", j >> 1); // arithmetical shift
Please let me know,
In case of nonnegative numbers, both kinds of right-shifts are the same. The difference appears only when the number to shift is negative.
Actually the C standard does not specify when should >>
perform logical or arithmetic shift when the number is negative, but typically, it performs arithmetic shift. To perform logical shift, the number must be cast to the corresponding unsigned type, for example:
int x = -2;
int y = x >> 1; // arithmetic shift.
assert (y == -1);
int z = (unsigned)x >> 1; // logical shift.
assert (z == 0x7FFFFFFF);
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