I want to check if the LSB is 0.
if(some_size_t & 1){} works fine
But why is if(some_size_t & 0){//This parts is unreachable} never reachable?
Because in order to ever get the value 1 i.e. true both operands for the logical and bitwise & i.e. AND operator have to be 1. Its so called truth table is
op1 | op2 | op1 AND op2
=====================
0 | 0 | 0
1 | 0 | 0
0 | 1 | 0
1 | 1 | 1
Because your value, e.g. op2, 0 has only zeros, no ones, you will always get only zeros as a result, no matter the other operand. And 0 will evaluate to false. It's what we call a contradiction in logic, often noted with a up side down T or \bot in latex.
As then the if condition is always false, the code in its body will never be executed, i.e. is unreachable.
Related
I have a question: Is bitwise anding transitive, particularly in C and C++?
Say res=(1 & 2 & 3 & 4), is this same as res1=(1&2) and res2=(3&4) and
res= (res1 & res2). Will this be same?
Yes, bitwise AND is transitive as you've used the term.
It's perhaps easier to think of things as a stack of bits. So if we have four 4-bit numbers, we can do something like this:
A = 0xB;
B = 0x3;
C = 0x1;
D = 0xf;
If we simply stack them up:
A 1 0 1 1
B 0 0 1 1
C 0 0 0 1
D 1 1 1 1
Then the result of a bitwise AND looks at one column at a time, and produces a 1 for that column if and only if there's a 1 for every line in that column, so in the case above, we get: 0 0 0 1, because the last column is the only one that's all ones.
If we split that in half to get:
A 1 0 1 1
B 0 0 1 1
A&B 0 0 1 1
And:
C 0 0 0 1
D 1 1 1 1
C&D 0 0 0 1
Then and those intermediate results:
A&B 0 0 1 1
C&D 0 0 0 1
End 0 0 0 1
Our result is still going to be the same--anywhere there's a zero in a column, that'll produce a zero in the intermediate result, which will produce a zero in the final result.
The term you're looking for is associative. We generally wouldn't call a binary operator "transitive". And yes, & and | are both associative, by default. Obviously, you could overload the operators to be something nonsensical, but the default implementations will be associative. To see this, consider one-bit values a, b, and c and note that
(a & b) & c == a & (b & c)
because both will be 1 if and only if all three inputs are 1. And this is the operation that is being applied pointwise to each bit in your integer values. The same is true of |, simply replacing 1 with 0.
There are also some issues to consider if your integers are signed, as the behavior is dependent on the underlying bit representation.
when x = 1
what should
! x | x
supposed to be?
I am really confused as what I did it:
x = 1 = 01 in binary,
!x = 10
!x | x = 11 = 3in decimal.
But it should be 1. (Even try going hexadecimal (something lengthy but what I am learning as an undergrad), I got -1)
The ! is the logical negation operator. If you give it nonzero things it gives you back a zero. If you give it a zero it gives you back 1.
So
!x | x
=> !1 | 1
=> 0 | 1
=> 1
Note how this is different from the bitwise negation operator, ~. If you had used this instead of !, things would have worked out like this (assuming 8-bit values, you can scale up to 32 or 64 or whatever):
~x | x
=> ~1 | 1
=> 11111110 | 00000001
=> 11111111
=> -1
It all comes down to understanding the difference between ! and ~. It's not obvious; it's just something you have to get used to. Just as a reminder
!7 = !23423523 = !46 = !(-200) = !1 = 0
For any nonzero x, !x = 0. That's just how it is. The idea behind that is C takes 0 as false and anything else as true. So, since a value like 70343 or 1 counts as true, applying ! to it gives false, or 0.
the ! operator is for logical negation, !x is equivalent to x == 0 for both numeric and pointer types.
!x has type int and a value of 1 if x compares equal to 0 and has a value of 0 for all other cases.
I have been going over a perl book i have recently purchased, and while reading I noticed a block of code that confused me..
use integer;
$value = 257;
while($value){
unshift #digits, (0..9,a..f)[$value & 15];
$value /= 16;
}
print digits;
the book mentions the purpose was to reverse the order of digits. however, new to perl I am having trouble figuring out what [$value & 15] is doing.
It's a bitwise and operation.
What it's doing is performing a bitwise and using the value of 15 and whatever value is contained in $value.
The resulting value is the decimal value that corresponds to the result of a bitwise and with the lower 4 bits of the value.
Ex:
$value = 21
which has a binary representation of: 0b10101
Performing a bitwise and with 15 means that any bits in $value will be zeroed if they are either outside the lower 4 bit range, or contain no 1's in the lower 4 bits.
The result is:
0b10101
&
0b 1111
-------
0b00101 = 5
Looking up the truth tables for performing bitwise operations will help with stuff like this in the future, but when performing an AND with any value, the result is only true, when both bits are 1, 0 otherwise.
V1 | V2 | V1 & V2
-----------------
0 | 0 | 0
0 | 1 | 0
1 | 0 | 0
1 | 1 | 1
What operation does the following ‘C’ statement perform?
star = star ^ 0b00100100;
(A) Toggles bits 2 and 5 of the variable star.
(B) Clears all bits except bits 2 and 5 of the variable star.
(C) Sets all bits except bits 2 and 5 of the variable star.
(D) Multiplies value in the variable star with 0b00100100.
I'm still clueless about this. Can someone help me out?
XOR operator (also called "logical addition") is defined like this:
a b a^b
-----------
0 0 0
0 1 1
1 0 1
1 1 0
So a^0 leaves a intact while a^1 toggles it.
For multiple-bit values, the operation is performed bitwise, i.e. between corresponding bits of the operands.
If you know how XOR works, and you know that ^ is XOR in C, then this should be pretty simple. You should know that XOR will flip bits where 1 is set, bits 2 and 5 of 0b00100100 are set, therefore it will flip those bits.
From an "during the test" standpoint, let's say you need to prove this to yourself, you really don't need to know the initial value of star to answer the question, If you know how ^ works then just throw anything in there:
00100100
^10101010 (star's made up value)
---------
10001110 (star's new value)
bit position: | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0
|---|---|---|---|---|---|---|---
star's new v: | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0
|---|---|---|---|---|---|---|---
star's old v: | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0
Then check your answers again, did it:
(A) Toggles bits 2 and 5 of the variable star. (Yes)
(B) Clears all bits except bits 2 and 5 of the variable star. (Nope)
(C) Sets all bits except bits 2 and 5 of the variable star. (Nope)
(D) Multiplies value in the variable star with 0b00100100. (36x170 = 142? Nope)
It is (A) toggles bits 2 and 5.
The following is the truth table for the XOR operation:
x y x^y
0 0 0
1 0 1
0 1 1
1 1 0
You can see from the table that x XOR 0 = x and x XOR 1 = !x.
XOR is a bitwise operation, so it operates on individual bits. Therefore if you XOR star with some constant, it will toggle the 1 bits in the constant.
You can find some explanation e.g. here.
The exclusive OR has this truth table:
A B A^B
-----------
1 1 0
1 0 1
0 1 1
0 0 0
We can see that if B is true (1) then A is flipped (toggled), and if it's false (0) A is left alone. So the answer is (A).
XOR operator returns 0 if both inputs are same otherwise returns 1 if both inputs are different.For Example the Given Truth Table :-
a=1 b=1 => a^b=0,
a=0 b=0 => a^b=0,
a=0 b=1 => a^b=1,
a=1 b=0 => a^b=1.
well xor is binary operator that work on bits of 2 nos.
rule of xoring:for same bit ans is 0 and for different bit ans is 1
let
a= 1 0 1 0 1 1
b= 0 1 1 0 1 0
--------------
c= 1 1 0 0 0 1
--------------
compare bit of a and b bit by bit
if same put 0 else put 1
xor is basically used to find the unique in given set of duplicate no.
just xor all nos. and u will get the unique one(if only single unique is present)
Is there an XOR operator or equivalent function in SQL Server (T-SQL)?
There is a bitwise XOR operator - the caret (^), i.e. for:
SELECT 170 ^ 75
The result is 225.
For logical XOR, use the ANY keyword and NOT ALL, i.e.
WHERE 5 > ANY (SELECT foo) AND NOT (5 > ALL (SELECT foo))
Using boolean algebra, it is easy to show that:
A xor B = (not A and B) or (A and not B)
A B | f = notA and B | g = A and notB | f or g | A xor B
----+----------------+----------------+--------+--------
0 0 | 0 | 0 | 0 | 0
0 1 | 1 | 0 | 1 | 1
1 0 | 0 | 1 | 1 | 1
1 1 | 0 | 0 | 0 | 0
As clarified in your comment, Spacemoses, you stated an example: WHERE (Note is null) ^ (ID is null). I do not see why you chose to accept any answer given here as answering that. If i needed an xor for that, i think i'd have to use the AND/OR equivalent logic:
WHERE (Note is null and ID is not null) OR (Note is not null and ID is null)
That is equivalent to:
WHERE (Note is null) XOR (ID is null)
when 'XOR' is not available.
MS SQL only short form (since SQL Server 2012):
1=iif( a=b ,1,0)^iif( c=d ,1,0)
The xor operator is ^
For example: SELECT A ^ B where A and B are integer category data types.
It is ^ http://msdn.microsoft.com/en-us/library/ms190277.aspx
See also some code here in the middle of the page How to flip a bit in SQL Server by using the Bitwise NOT operator
<> is generally a good replacement for XOR wherever it can apply to booleans.
From your comment:
Example: WHERE (Note is null) ^ (ID is null)
you could probably try:
where
(case when Note is null then 1 else 0 end)
<>(case when ID is null then 1 else 0 end)