I am currently learning C++. I was trying to compute power of an integer using the expression:
val=10^1;
Instead of expected answer 10, the result was 11. I have fixed the problem by using pow function of math.h library but I am wondering why this statement is giving me the wrong result.
No! Do you think that is the power? Don't forgot this (In C++ and some of the programming languages):
Be sure to read this:
Operators (cplusplus)
Bitwise operation (wikipedia)
A bitwise XOR takes two bit patterns of equal length and performs the logical exclusive OR operation on each pair of corresponding bits. The result in each position is 1 if only the first bit is 1 or only the second bit is 1, but will be 0 if both are 0 or both are 1. In this we perform the comparison of two bits, being 1 if the two bits are different, and 0 if they are the same. For example:
0101 (decimal 5)
XOR 0011 (decimal 3)
= 0110 (decimal 6)
The bitwise XOR may be used to invert selected bits in a register (also called toggle or flip). Any bit may be toggled by XORing it with 1. For example, given the bit pattern 0010 (decimal 2) the second and fourth bits may be toggled by a bitwise XOR with a bit pattern containing 1 in the second and fourth positions:
0010 (decimal 2)
XOR 1010 (decimal 10)
= 1000 (decimal 8)
This technique may be used to manipulate bit patterns representing sets of Boolean states.
Source: Wikipedia
That's the bitwise exclusive-or operator, not power. In binary:
10 = 1010
1 = 0001
val= 1011 = 11 in decimal
In C and C++, 10^1 is 10 XOR 1, not 10 to the power of 1.
^ is the binary XOR operator in C++:
10 ^ 1 =
00001010
^ 00000001
= --------
00001011 = 11
Because ^ is the exclusive or operator and not the exponentiation operator. Basically, because the last bit of 10 in binary is 0, by applying exclusive or of 1 the last bit gets converted to 1 because it is different than 0.
Because in C++ there is no power operator: ^ is a XOR.
1010 is 10102 in binary; 110 is 00012. XORing them together gives 10112, which is 1110.
If you would like to obtain 10n, use the pow function from <cmath> header:
val=pow(10, 1);
Related
I came across a part of code that I cannot understand.
for (unsigned int i = (x & 0b1); i < x; i+= 2)
{
// body
}
Here, x is from 0 to 5.
What is meant by 0b1? and what would be the answers for eg: (0 & 0b1), (4 & 0b1) etc?
0b... is a binary number, just like 0x... is hex and 0... is octal.
Thus 0b1 is same as 1.
1b0 is illegal, the first digit in those must always be 0.
As previous answers said, it is the binary representation of the integer number 1, but they don't seem to have fully answered your question. This has a lot of layers so I'll briefly explain each.
In this context, the ampersand is working as a bitwise AND operator. i & 0b1 is (sometimes) a faster way of checking if an integer is even as opposed to i % 2 == 0.
Say you have int x = 5 and you'd like to check if it's even using bitwise AND.
In binary, 5 would be represented as 0101. That final 1 actually represents the number 1, and in binary integers it's only present in odd numbers. Let's apply the bitwise AND operator to 5 and 1;
0101
0001
&----
0001
The operator is checking each column, and if both rows are 1, that column of the result will be 1 – otherwise, it will be 0. So, the result (converted back to base10) is 1. Now let's try with an even number. 4 = 0100.
0100
0001
&----
0000
The result is now equal to 0. These rules apply to every single integer no matter its size.
The higher-level layer here is that in C, there is no boolean datatype, so booleans are represented as integers of either 0 (false) or any other value (true). This allows for some tricky shorthand, so the conditional if(x & 0b1) will only run if x is odd, because odd & 0b1 will always equal 1 (true), but even & 0b1 will always equal 0 (false).
I'm new to x64-64, just a question on how does CF get set? I was reading a textbook which says:
CF: Carry flag is used when most recent operation generated a carry out of the most significant bit. Used to detect overflow for unsigned operations.
I have two questions:
Q1-suppose we used one of the add instructions to perform the equivalent of the C assignment t = a+b, where variables a, b, and t are integers (only 3 bits for simplicity), so for 011(a) + 101(b) = 1000 = 000, since we have a carry out bit 1 in the fourth digit, so CF flag will be set to 1, is my understanding correct?
Q2-if my understanding in Q1 is true, and suppose we used one of the sub instructions to perform the equivalent of the C assignment t = a-b, where a, b, and t are unsigned integers, since a, b are unsigned, we can't actually do a+(-b), and I don't get how we can make 011(a) - 101(b) carry out of the most significant bit?
The carry flag is often called "borrow" when performing a subtraction. After a subtraction, it set if a 1 had to be borrowed from the next bit (or would have been borrowed if you used the grade-school subtraction method). The borrow flag is like a -1 in that bit position:
011 -1 211
- 101 -> - 101
----- -----
B 110
You can get the same result by adding a zero to the arguments, and then the carry or borrow will be the high bit of the result:
0011 - 0101 = 0011 + (-0101) = 0011 + 1011 = 1110
I am new to Bitwise Operator,I found the following line in wikipedia of "Bitwise operator" -(https://en.wikipedia.org/wiki/Bitwise_operation )
If the binary number is treated as ones' complement, then the same
right-shift operation results in division by 2n and rounding toward
zero.
But not sure what it meant.Please help me to understand this.
It's about Right Shift Operation. Below answer is taken from my own blog:
Bitwise Operations
>> is a right shift bitwise operator which works exactly like left shift operator but it’s different in 2 aspects only.
If number is negative (most significant bit is 1) , it will fill gaps with 1 otherwise with 0
It shifts bits from right
7 >> 2
0000 0111 >> 2
0000 01 (See two 1’s from right have been shifted)
00 0000 01 (Added two 0’s, because number is positive)
0000 0001 = 1
So 7 >> 2 = 1
For negative number, it fills gaps with 1 instead of 0.
Remember 7 >> 2 = 1 and -7 >> 2 = -1.
It just preserves the sign, everything else remains same. If you closely observe right shift operator, you will find that it can be written as:
7 >> 2 => (7) / (2^2) => 7/4 => 1
This is what Wikipedia means, you can take right shift operator as:
NUMBER (7) / power (2, NUMBER_OF_TIMES (2)) and rounds it off
This question already has answers here:
bit-wise operation unary ~ (invert)
(5 answers)
Closed 7 years ago.
int x=10;
cout<<~x;
this code prints -11. if it was simple inversion then for 00001010 the bits should be 11110101, which on conversion to decimal is -117. I have tried searching but no luck pls tell what is happening here?
I am using mingw compiler, if its of any help.
That is working as expected. "11110101" is -11 in two's complement.
As a side note, "int" is either 16 or 32 bits, so you're actually talking about
"00000000 00001010" or "00000000 00000000 00000000 00001010" respectively.
~x is equal to −x − 1. Therefore, ~10 = -10 - 1 = -11.
Using 1-byte , 10 is represented as 0000 1010. Its bit-wise NOT is 1111 0101. Generally computer represents signed integers in 2's complement format. So, decimal equivalent of 1111 0101 is -11, How?
An N-bit number w represented in 2's complement as aN-1aN-2....a0 can be converted to decimal as
Therefore,
1111 01012 = -1*27 1*26 + 1*25 + 1*24 + 0*23 + 1*22 + 0*21 + 1*20 = -128 + 117 = -11
The ~ operator works as you think it does. It's just negative numbers don't work how you think they do. Negative numbers are encoded as a two's complement http://en.wikipedia.org/wiki/Two%27s_complement
Basically, the value of the highest bit is subtracted from the value of the lower bits.
I'd like to confirm whether I grasped the concept of CRC calculations correctly. I'll provide two examples, the first is calculating the remainder using normal subtraction, the second uses this weird XOR stuff.
Data bits: D = 1010101010.
Generator bits: G = 10001.
1) Subtraction approach to calculate remainder:
10101010100000
10001|||||||||
-----|||||||||
10001|||||||
10001|||||||
-----|||||||
000000100000
10001
-----
1111
R = 1111.
2) XOR approach:
10101010100000
10001|||||||||
-----|||||||||
10001|||||||
10001|||||||
-----|||||||
00000010000|
10001|
------
000010
R = 0010.
Appending 1111 at the end does not satisfy the need since
10927 % 17 != 0
.
Note that as per the definition, the division should be modulo division as it is based upon modulo mathematics.
Both answers are correct. =)
(To recheck the first answer:
10101010100000 (binary) mod 10001 (binary)
= 10912 (decimal) mod 17 (decimal)
= 15 (decimal)
= 1111 (binary).)
Subtraction is wrongly done. In binary modulo, subtraction, addition, division, and multiplication are the same. So, XOR is correct.