I don't know why this code works. it supposed to print out every student which follows chem. but why does a number for instance 21&4(student 123001) evaluate to true while a number like 49&4(123008) doesn't?
I think it is due to bit operation AND.
In binary
49 is 110001
4 is 000100
& = 000000
So it evaluates to false
wheras
21 is 10101
4 is 00100
& = 00100
So you get a non-zero result which is true.
Related
I'm trying to understand how a binary addition and logical OR table differs.
does both carry forward 1 or if not which one does carry forward operation and which does not?
The exclusive-or (XOR) operation is like binary addition, except that
there is no carry from one bit position to the next. Thus, each bit
position can be evaluated independently of the rest.
I'll attempt to clarify a few points with a few illustrations.
First, addition. Basically like adding numbers in grade school. But if you have a 1-bit aligned with a 1-bit, you get a 0 with a 1 carry (i.e. 10, essentially analogous to 5 plus 5 in base-10). Otherwise, add them like 'regular' (base-10) numbers. For instance:
₁₁₁
1001
+ 1111
______
11000
Note that in the left-most column two 1's are added to give 10, which with another 1 gives 11 (similar to 5 + 5 + 5).
Now, assuming by "logical OR" you mean something along the lines of bitwise OR (an operation which basically performs the logical OR (inclusive) operation on each pair of corresponding bits), then you have this:
1001
| 1111
______
1111
Only case here you should have a 0 bit is if both bits are 0.
Finally, since you tagged this question xor, which I assume is bitwise as well.
1001
^ 1111
______
0110 = 110₂
In this case, two 1-bits give a 0, and of course two 0-bits give 0.
With a logical OR you get a logical result (Boolean). IOW true OR true is true (anything other than false OR false is true). In some languages (like C) any numeric value other than 0 means true. And some languages use an explicit datatype for true, false (bool, Boolean).
In case of binary OR, you are ORing the bits of two binary values. ie: 1 (which is binary 1) bitwise OR 2 (which is binary 10) is binary 11:
01
10
11
which is 3. Thus binary OR is also an addition when the values do not have shared bits (like flag values).
Please, explain why by doing as follows I'll get a second bit of the number stored in i in it's internal representation.
(i & 2) / 2;
Doing i & 2 masks out all but the second bit in i. [1]
That means the expression evaluates to either 0 or 2 (binary 00 and 10 respectively).
Dividing that by 2 gives either 0 or 1 which is effectively the value of the second bit in i.
For example, if i = 7 i.e. 0111 in binary:
i & 2 gives 0010.
0010 is 2 in decimal.
2/2 gives 1 i.e. 0001.
[1] & is the bitwise AND in C. See here for an explanation on how bitwise AND works.
i & 2 masks out all but the second bit.
Dividing it by 2 is the same as shifting down 1 bit.
e.g.
i = 01100010
(i & 2) == (i & 00000010) = 00000010
(i & 2) / 2 == (i & 2) >> 1 = 00000001
The & operator is bitwise AND: for each bit, the result is 1 only if the corresponding bits of both arguments are 1. Since the only 1 bit in the number 2 is the second-lowest bit, a bitwise AND with 2 will force all the other bits to 0. The result of (i & 2) is either 2 if the second bit in i is set, or 0 otherwise.
Dividing by 2 just changes the result to 1 instead of 2 when the second bit of i is set. It isn't necessary if you're just concerned with whether the result is zero or nonzero.
2 is 10 in binary. & is a bitwise conjunction. So, i & 2 gets you the second-from-the-end bit of i. And dividing by 2 is the same as bit-shifting by 1 to the right, which gets the value of the last bit.
Actually, shifting to the right would be better here, as it clearly states your intent. So, this code would be normally written like this: (i & 0x02) >> 1
What does the following condition effectively check in C :
if(a & (1<<b))
I have been wracking my brains but I can't find a pattern.
Any help?
Also I have seen this used a lot in competitive programming, could anyone explain when and why this is used?
It is checking whether the bth bit of a is set.
1<<b will shift over a single set bit b times so that only one bit in the bth position is set.
Then the & will perform a bitwise and. Since we already know the only bit that is set in 1<<b, either it is set in a, in which case we get 1<<b, or it isn't, in which case we get 0.
In mathematical terms, this condition verifies if a's binary representation contains 2b. In terms of bits, this checks if b's bit of a is set to 1 (the number of the least significant bit is zero).
Recall that shifting 1 to the left by b positions produces a mask consisting of all zeros and a single 1 in position b counting from the right. A value of this mask is 2b.
When you perform a bitwise "AND" with such a mask, the result would be non-zero if, and only if, a's binary representation contains 2b.
Lets say for example a = 12 (binary: 1100) and you want to check that the third bit (binaries are read from right to left) is set to 1, to do that you can use & bitwise operator which work as following:
1 & 0 = 0
0 & 1 = 0
0 & 0 = 0
1 & 1 = 1
To check if the third bit in a is set to 1 we can do:
1100
0100 &
------
0100 (4 in decimal) True
if a = 8 (binary: l000) on the other hand:
1000
0100 &
------
0000 (0 in decimal) False
Now to get the 0100 value we can right shift 1 by 2 (1 << 2) wich will append two zeros from the right and we'll get 100, in binaries left trailing zeros doesn't change the value so 100 is the same as 0100.
I want to ask about C operator from this code. My friends ask it, but I never seen this operator:
binfo_out.biSizeImage = ( ( ( (binfo_out.biWidth * binfo_out.biBitCount) + 31) & ~31) / 8) * abs(out_bi.biHeight);
What this operator & ~31 mean? anybody can explain this?
The & operator is a bitwise AND. The ~ operator is a bitwise NOT (i.e. inverts the bits). As 31 is binary 11111, ~31 is binary 1111111....111100000 (i.e. a number which is all ones, but has five zeroes at the end). Anding a number with this thus clears the least significant five bits, which (if you think about it) rounds down to a multiple of 32.
What does the whole thing do? Note it adds 31 first. This has the effect that the whole thing rounds something UP to the next multiple of 32.
This might be used to calculate (for instance), how many bits are going to be used to store something if you can only use 32 bit quantities to store them, as there is going to be some wastage in the last 32 bit number.
31 in binary representation will be 11111 so ~31 = 5 zeros 00000 preceeded by 1's. so it is to make last 5 bits zero. i.e. to mask the last 5 bits.
here ~ is NOT operator i.e. it gives 1's complement. and & is AND operator.
& is the bitwise AND operator. It and's every corresponding bit of two operands on its both sides. In an example, it does the following:
Let char be a type of 8 bits.
unsigned char a = 5;
unsigned char b = 12;
Their bit representation would be as follows:
a --> 0 0 0 0 0 1 0 1 // 5
b --> 0 0 0 0 1 1 0 0 // 12
And the bitwise AND of those would be:
a & b --> 0 0 0 0 0 1 0 0 // 8
Now, the ~ is the bitwise NOT operator, and it negates every single bit of the operand it prefixes. In an example, it does the following:
With the same a from the previous example, the ~a would be:
~a --> 1 1 1 1 1 0 1 0 // 250
Now with all this knowledge, x & ~31 would be the bitwise AND of x and ~31, where the bit representation of ~31 looks like this:
~31 --> 1111 1111 1111 1111 1111 1111 1110 0000 // -32 on my end
So the result would be whatever the x has on its bits, other than its last 5 bits.
& ~31
means bitwise and of the operand on the left of & and a bitwise not of 31.
http://en.wikipedia.org/wiki/Bitwise_operation
The number 31 in binary is 11111 and ~ in this case is the unare one's compliment operator. So assuming 4-byte int:
~31 = 11111111 11111111 11111111 11100000
The & is the bitwise AND operator. So you're taking the value of:
((out_bi.biWidth * out_bi.biBitCount) + 31)
And performing a bitwise AND with the above value, which is essentially blanking the 5 low-order bits of the left-hand result.
I came across a common programming interview problem: given a list of unsigned integers, find the one integer which occurs an odd number of times in the list. For example, if given the list:
{2,3,5,2,5,5,3}
the solution would be the integer 5 since it occurs 3 times in the list while the other integers occur even number of times.
My original solution involved setting up a sorted array, then iterating through the array: For each odd element I would add the integer, while for each even element I would subtract; the end sum was the solution as the other integers would cancel out.
However, I discovered that a more efficient solution existed by simply performing an XOR on each element -- you don't even need a sorted array! That is to say:
2^3^5^2^5^5^3 = 5
I recall from a Discrete Structures class I took that the Associate Property is applicable to the XOR operation, and that's why this solution works:
a^a = 0
and:
a^a^a = a
Though I remember that the Associative Property works for XOR, I'm having trouble finding a logical proof for this property specific to XOR (most logic proofs on the Internet seem more focused on the AND and OR operations). Does anyone know why the Associative Property applies to the XOR operation?
I suspect it involves an XOR identity containing AND and/or OR.
The associative property says that (a^b)^c = a^(b^c). Since XOR is bitwise (the bits in the numbers are treated in parallel), we merely need to consider XOR for a single bit. Then the proof can be done by examining all possibilities:
abc (a^b) (a^b)^c (b^c) a^(b^c)
000 0 0 0 0
001 0 1 1 1
010 1 1 1 1
011 1 0 0 0
100 1 1 0 1
101 1 0 1 0
110 0 0 1 0
111 0 1 0 1
Since the third column, (a^b)^c, is identical to the fifth column, a^(b^c), the associative property holds.
As long as a ^ b == ~a & b | a & ~b, you can proove that :
(a ^ b) ^ c = ~((~a & b) | (a & ~b)) & c | ((~a & b) | (a & ~b)) & ~c
and
a ^ (b ^ c) = a & ~((~b & c) | (b & ~c)) | ~a & ((~b & c) | (b & ~c))
Are equals.