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I need a really short way to extract the last 4 digits of a hex number. So an input of 0x2479c should throw an output of 0x479c. I want to avoid converting and reconverting to binary.
Modulo division, which would generally work for decimal numbers, does not work in this case.
0x2479c modulo 0xffff = 0x479e
which isn't correct. I'm trying to achieve this is c/c++.
You should use a mask and then a byte wise 'and' with an other value. In your case :
0x2479c & 0x0ffff
Either use a mask
0x2479c & 0x0ffff
or the modulo operator
0x2479c % (0x10000);
You were off by one in the operand of modulo.
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How can I count the even and the odd integers in a given number in C++?
For example: The user inputs: 32478
output: 3 even numbers and 2 odd numbers.
The basic algorithm is:
Take the number modulo 2 (num % 2). If the result is 1 then the number is odd; increment the odd counter. If not then it's even; increment the even counter.
Divide the number by 10, dropping the remainder. (num /= 10)
Go back to step 1 if the number isn't zero.
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How do I design an algorithm that takes two floats and multiplies them using only addition, bit shifting and bitwise operations?
I have already found one like this for integers, but that doesn't work with floats.
I have also found another that is much like what I need but log is also prohibited in my case.
The floats are stored according to the IEEE754 standard. I have also tried to keep their exponent part, and bitwise multiply their fractional part with no luck.
According to http://en.wikipedia.org/wiki/IEEE_floating_point, an IEEE754 number x = (-1)^s * c * b^q is represented by s,c,b,q , all are integers. for Two floating point numbers with the same base b is the same.
So the multiplication of two floating point numbers x and y is:
(-1)^(s1+s2)*c1*c2*b^(q1+q2) so the new floating point is represented by: s1+s2, c1*c2, b q1+q2 so you only have left to deal with multiplication of c1 and c2, both are integers so you are done.
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How can I write a Reg-ex Expression to check whether a string is a binary multiple of 4? I am not good at making DFA and finding expressions.
A multiple of 4 in binary is any binary number that ends with 00, so this regexp should do it:
^(?:[10]*00|00?)$
If you mean a multiple of 4 in decimal, I wouldn't do that with a regexp, except perhaps to verify that it's a number. Then I'd parse it and check whether number % 4 is zero.
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Can a regex pattern enforce the following combination of constraints on a numeric value?
the number must be >= 1 and <= 999
(decimal point cannot be the first character in the string?)
the number can be an integer or a number with a fractional component
when it has a fractional component,
no more than 2 digits to the right of the decimal point
EDIT: but at least one digit to the right
must not have leading zero(s)
Perl syntax:
^+?(?:(?:999(?:\.0{1,2})?)|(?:(?!999)[1-9]\d{0,2}(?:\.\d{1,2})?))$
I'll go with this one:
/^
(?:
999.0* # 999.000
| # or
[1-9]\d{,2}((?<!999)\.\d+)? # 1-998 plus optional .\d*
)
$/x
Yes..
^(?!999[.](0*[1-9]+$))(?!0|[.])\d{1,3}([.]\d{1,2})?$
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I need an regular expression pattern to only accept positive whole numbers. It can also accept a single zero.
I do not want to accept decimals, negative number and numbers with leading zeros.
Any suggestions?
^(0|[1-9][0-9]*)$
"[1-9][0-9]*|0"
I'd just use "[0-9]+" to represent positive whole numbers.
This will allow decimal numbers (or whole numbers) that don't start with zero:
^(([1-9]*)|(([1-9]*)\.([0-9]*)))$
If you want to allow numbers that start with zero, you can do :
^(([0-9]*)|(([0-9]*)\.([0-9]*)))$
/([1-9][0-9]*)|0/
/^0|[1-9]\d*$/