For integers, I can use 'BigInt' if it throws a similar error, what about hexadecimal values? I strictly don't want to convert the hex value into an integer and then use it. Can somebody help?
val a = 0x1265465678687564534344536
<console>:1: error: integer number too large
That class BigInteger has a constructor that takes a String argument, so the simple answer is: instead of using a numeric literal, that is out of any meaningful range, create a BigInteger based on "0x1265465678687564534344536" instead!
Use the BigInteger class that takes in a string parameter and pass the radix as 16 to indicate hexadecimal string.
new BigInteger(string, 16)
Related
I am learning dart programming language for Flutter. In the integer class what does the word radix means ? Please explain me this. Thanks
Sometimes we have to work with string in radix number format. Dart int parse() method also supports convert string into a number with radix in the range 2..36:
For example, we convert a Hex string into int:
var n_16 = int.parse('FF', radix: 16);
The output of the code = 255
Using radix function we can also convert Binary numbers into decimal numbers like this
var decimal = int.parse('1001001', radix:2)'
I deal with numbers of this form 1.446267186999E7 and i want to represent them without E.
For example 1.446267186999E7 i want it to be 14462671.86999 .
How do i convert it to this form without getting the :
error: integer number too large.
Thanks for the helpers.
Try this:
BigDecimal(1.446267186999E7).toString
The BigDecimal.toString method will give you the string representation of the number in decimal form.
That is just a formatting problem if you store it as a double.
import java.text.DecimalFormat
val d: Double = 1.446267186999E7
val decimalFormat: DecimalFormat = new DecimalFormat("0.#####")
println(decimalFormat.format(d))
should give you 14462671.86999
You probably after BigDecimal. In terms of string formatting, look at the .format method, or printf
Could someone explain why I can not use int() to convert an integer number represented in string-scientific notation into a python int?
For example this does not work:
print int('1e1')
But this does:
print int(float('1e1'))
print int(1e1) # Works
Why does int not recognise the string as an integer? Surely its as simple as checking the sign of the exponent?
Behind the scenes a scientific number notation is always represented as a float internally. The reason is the varying number range as an integer only maps to a fixed value range, let's say 2^32 values. The scientific representation is similar to the floating representation with significant and exponent. Further details you can lookup in https://en.wikipedia.org/wiki/Floating_point.
You cannot cast a scientific number representation as string to integer directly.
print int(1e1) # Works
Works because 1e1 as a number is already a float.
>>> type(1e1)
<type 'float'>
Back to your question: We want to get an integer from float or scientific string. Details: https://docs.python.org/2/reference/lexical_analysis.html#integers
>>> int("13.37")
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 10: '13.37'
For float or scientific representations you have to use the intermediate step over float.
Very Simple Solution
print(int(float(1e1)))
Steps:-
1- First you convert Scientific value to float.
2- Convert that float value to int .
3- Great you are able to get finally int data type.
Enjoy.
Because in Python (at least in 2.x since I do not use Python 3.x), int() behaves differently on strings and numeric values. If you input a string, then python will try to parse it to base 10 int
int ("077")
>> 77
But if you input a valid numeric value, then python will interpret it according to its base and type and convert it to base 10 int. then python will first interperet 077 as base 8 and convert it to base 10 then int() will jsut display it.
int (077) # Leading 0 defines a base 8 number.
>> 63
077
>> 63
So, int('1e1') will try to parse 1e1 as a base 10 string and will throw ValueError. But 1e1 is a numeric value (mathematical expression):
1e1
>> 10.0
So int will handle it as a numeric value and handle it as though, converting it to float(10.0) and then parse it to int. So Python will first interpret 1e1 since it was a numric value and evaluate 10.0 and int() will convert it to integer.
So calling int() with a string value, you must be sure that string is a valid base 10 integer value.
int(float(1e+001)) will work.
Whereas like what others had mention 1e1 is already a float.
I want to convert the text hello to ascii decimal in PARI/GP. After that I will
concatenate the values.
I initialize a Vecsmall(hello), after that I run a loop to concatenate the ascii decimal values,
I want to use this concatenated value to * by certain values. The value is now in String type, In Java, there is a Integer.parseInt() to convert the string to int. I wonder if there is a similar function in PARI/GP?
v=Vecsmall("hello");'
for (i = 1, length(v), text=Str(text,v[i]););
//is there any similar function like Integer.praseInt(text) in PARI?
You can use eval
eval(text)
or else a combination of Vecsmall and fromdigits which is faster:
fromdigits(apply(n->n-49, Vec(text)))
Does anyone have a code snippet or a class that will take a long long and turn it into a 16 byte Hex string?
I'm looking to turn data like this
long long decimalRepresentation = 1719886131591410351;
and turn it into this
//Base 16 Hex Output: 17DE435307A07300
The %x operator doesn't want to work for me
NSLog(#"Hex: %x",decimalRepresentation);
//console : "Hex: 7a072af"
As you can see that's not even close. Any help is truly appreciated!
%x prints an unsigned integer in hexadecimal representation and sizeof(long long) != sizeof(unsigned). See e.g. "Data Type Size and Alignment" in the 64bit transitioning guide.
Use the ll specifier (thats two lower-case L) to get the desired output:
NSLog(#"%llx", myLongLong);