There is actually a (subtle) difference between the two. Imagine you have the following code in File1.cs:
// File1.cs
using System;
namespace Outer.Inner
{
class Foo
{
static void Bar()
{
double d = Math.PI;
}
}
}
Now imagine that someone adds another file (File2.cs) to the project that looks like this:
// File2.cs
namespace Outer
{
class Math
{
}
}
The compiler searches Outer
before looking at those using
directives outside the namespace, so it finds Outer.Math
instead of System.Math
. Unfortunately (or perhaps fortunately?), Outer.Math
has no PI
member, so File1 is now broken.
This changes if you put the using
inside your namespace declaration, as follows:
// File1b.cs
namespace Outer.Inner
{
using System;
class Foo
{
static void Bar()
{
double d = Math.PI;
}
}
}
Now the compiler searches System
before searching Outer
, finds System.Math
, and all is well.
Some would argue that Math
might be a bad name for a user-defined class, since there's already one in System
; the point here is just that there is a difference, and it affects the maintainability of your code.
It's also interesting to note what happens if Foo
is in namespace Outer
, rather than Outer.Inner
. In that case, adding Outer.Math
in File2 breaks File1 regardless of where the using
goes. This implies that the compiler searches the innermost enclosing namespace before it looks at any using
directive.
The bit shifting operators do exactly what their name implies. They shift bits. Here's a brief (or not-so-brief) introduction to the different shift operators.
The Operators
>>
is the arithmetic (or signed) right shift operator.
>>>
is the logical (or unsigned) right shift operator.
<<
is the left shift operator, and meets the needs of both logical and arithmetic shifts.
All of these operators can be applied to integer values (int
, long
, possibly short
and byte
or char
). In some languages, applying the shift operators to any datatype smaller than int
automatically resizes the operand to be an int
.
Note that <<<
is not an operator, because it would be redundant.
Also note that C and C++ do not distinguish between the right shift operators. They provide only the >>
operator, and the right-shifting behavior is implementation defined for signed types. The rest of the answer uses the C# / Java operators.
(In all mainstream C and C++ implementations including GCC and Clang/LLVM, >>
on signed types is arithmetic. Some code assumes this, but it isn't something the standard guarantees. It's not undefined, though; the standard requires implementations to define it one way or another. However, left shifts of negative signed numbers is undefined behaviour (signed integer overflow). So unless you need arithmetic right shift, it's usually a good idea to do your bit-shifting with unsigned types.)
Left shift (<<)
Integers are stored, in memory, as a series of bits. For example, the number 6 stored as a 32-bit int
would be:
00000000 00000000 00000000 00000110
Shifting this bit pattern to the left one position (6 << 1
) would result in the number 12:
00000000 00000000 00000000 00001100
As you can see, the digits have shifted to the left by one position, and the last digit on the right is filled with a zero. You might also note that shifting left is equivalent to multiplication by powers of 2. So 6 << 1
is equivalent to 6 * 2
, and 6 << 3
is equivalent to 6 * 8
. A good optimizing compiler will replace multiplications with shifts when possible.
Non-circular shifting
Please note that these are not circular shifts. Shifting this value to the left by one position (3,758,096,384 << 1
):
11100000 00000000 00000000 00000000
results in 3,221,225,472:
11000000 00000000 00000000 00000000
The digit that gets shifted "off the end" is lost. It does not wrap around.
Logical right shift (>>>)
A logical right shift is the converse to the left shift. Rather than moving bits to the left, they simply move to the right. For example, shifting the number 12:
00000000 00000000 00000000 00001100
to the right by one position (12 >>> 1
) will get back our original 6:
00000000 00000000 00000000 00000110
So we see that shifting to the right is equivalent to division by powers of 2.
Lost bits are gone
However, a shift cannot reclaim "lost" bits. For example, if we shift this pattern:
00111000 00000000 00000000 00000110
to the left 4 positions (939,524,102 << 4
), we get 2,147,483,744:
10000000 00000000 00000000 01100000
and then shifting back ((939,524,102 << 4) >>> 4
) we get 134,217,734:
00001000 00000000 00000000 00000110
We cannot get back our original value once we have lost bits.
Arithmetic right shift (>>)
The arithmetic right shift is exactly like the logical right shift, except instead of padding with zero, it pads with the most significant bit. This is because the most significant bit is the sign bit, or the bit that distinguishes positive and negative numbers. By padding with the most significant bit, the arithmetic right shift is sign-preserving.
For example, if we interpret this bit pattern as a negative number:
10000000 00000000 00000000 01100000
we have the number -2,147,483,552. Shifting this to the right 4 positions with the arithmetic shift (-2,147,483,552 >> 4) would give us:
11111000 00000000 00000000 00000110
or the number -134,217,722.
So we see that we have preserved the sign of our negative numbers by using the arithmetic right shift, rather than the logical right shift. And once again, we see that we are performing division by powers of 2.
Best Answer
They didn't. You can do
&=
or|=
or^=
if you want.The
||
and&&
operators do not have a compound form because frankly, they're a bit silly. Under what circumstances would you want to saysuch that the right hand side is not evaluated if the left hand side does not change? It seems like only a few people would actually use this feature, so why put it in?
For more information about the compound operators, see my serious article here:
https://docs.microsoft.com/en-us/archive/blogs/ericlippert/compound-assignment-part-one
and the follow-up April-Fools article here:
https://docs.microsoft.com/en-us/archive/blogs/ericlippert/compound-assignment-part-two