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UInt62.max / 2 is represented by 0100..0000 in the memory. Add 1 and it will be 0100..0001. So, the first bit for the sign. And we take -1. But CPU thinks that it's -9 223 372 036 854 775 808. Why does it work so complexly?
You can see that it's truth because of the issue in the Swift playground: Why is UInt64 max equal -1 in Swift?
var max = UInt64.max / 2 + 1 // playground shows -1 because it treats it as Int64
Yes, in fact -1 is not represented as 1 with a sign bit, but rather as all bits set to one. This is called a "two's complement" representation, and is used in most of the modern processors.
Read more about it here:
https://en.wikipedia.org/wiki/Two%27s_complement
One of the reasons for that is that this way arithmetic operations that involve both negative and positive numbers are easier. If -1 was represented as 1 with a sign bit, and we attempted to add 1 to it in a naive way, we would get 2 with a sign bit instead of zero. With two's complement representation you can just add the numbers as if they were unsigned, and get the correct result.
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When i convert String values to Double, if the decimals are zeros (for example 10.0,11.0 etc), swift automatically rounds the number and removes the decimals. Is there a way to prevent swift from removing the zeros after the decimal point?
I gave it a try in the online compiler that repl.it provides. But it doesn't act the way you said. I wrote the following code:
let x = Double("10")!
print(x)
And the output I got was:
10.0
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I want to generate a random number with exactly 10-digits.
The random number cannot begin with zero, i.e. it must be a 10 digit number, not 10 random digits.
You need a random integer, between 1000000000 (the lowest integer with 10 digits) and 9999999999 (the highest number with 10 digits).
Note that 1000000000 = 1e9 and 9999999999 = 1e10 - 1
The random integer generation can easily achieved with randi (see the documentation here), giving it the correct minimum and maximum values...
n = randi([1e9, 1e10-1])
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Is there an efficient algorithm that determine the sum of two numbers, even if these numbers are greater than ten?
the user can only enter the numbers one digit at a time. Assuming only positive numbers can be entered the algorithm to parse it is as follows.
initialize accumulator to 0
for each digit the user enters
multiply the accumulator by 10
add the new digit to the accumulator
You will need to handle when the user enters the enter key to finish entering the number, and the above algorithm only works for numbers up to 255 (assuming 8 bit cells).
From then on you have a cell with a number. Do this again to get a cell with another number, then you can simply add them together normally.
Using http://fatiherikli.github.io/brainfuck-visualizer/ you can see the value of each cell numerically, not having to use all that dumb ASCII conversion stuff. Then, just program in your inputs and the algorithm.
++++++>+++++ 6 plus 5
[<+>-]
Adds one to 6 and subtracts one from 5 each iteration, making 11.
The visualizer will show that the cell is 11.
If you want to preserve the y value, add a "temp" cell
++++++>+++++>[-] #0: 6 #1: 5 #2: 0
<[<+>>+<-]
>[<+>-]
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Is there any opposite function of "primes" in matlab?
(Primes : Prime numbers less than or equal to input value)
What if i want to find greater than or equal to input value?
Thank you.
You always have to provide an upper limit. If you wanted prime numbers "greater than or equal to the given value," they would go to infinity.
I would suggest starting with "less than or equal to," provide the function with an upper limit, and then use a conditional to filter out those that are too small.
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When i try to display binary notations which start with a zero in the 1st bit position, matlab discards the zero and displays only the other 7 bits. How do I display the 1st position too?
ex: when i try to display "01101111", matlab displays it as "1101111", but I need the 1st bit position value also. Can some one please help.
In Matlab, to display the bit representation of a number you need to convert it into a string with dec2bin().
So, if you have x = 111, it's binary representation is:
dec2bin(111)
ans =
1101111
which retains only the significant bits. To force an 8-bit representation use:
dec2bin(111,8)
ans =
01101111
Note, how the result will be a string. If you want to retrieve bits in numeric format, then use bitget():
bitget(111,8:-1:1)
ans =
0 1 1 0 1 1 1 1
Basically, if your need is purely visual, use dec2bin2() otherwise for manipulating bits, use the bit-wise operations functions, which accept and return numeric types.