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In the first place, your approach to the problem is only in the grossest sense characterizable as converting decimal to octal. A better description would be that it is (trying to) convert decimal to the implementation's internal integer representation, which is likely characterizable as base-2, and / or base-2 8 and / or base-2 32.
I.e.: 01 is octal 1, 010 is octal 10, which is decimal 8, and 0 is octal 0 (which is decimal, and any other, 0). So yes, '0' is an octal. That's plain English translation of the grammar snippet in @Als's answer :-) An integer prefixed with 0x is not prefixed with 0. 0x is an explicitly different prefix.
You are safe up to 32767 (in octal: 77777), as 32768 (8*8*8*8*8 = 8^5 = (2^3)^5 = 2^15) is the first number, that requires six digits in octal: 100000. This o variable is not really needed, morever it will not work when int is signed 16-bit (on some ancient system), so from this point it's better to just print separate digits.
5. By "decimal integer" I hope you mean a string that uses decimal to represent an integer. Integer types, like int, do not have a base. Or if you insist that they must have a base because of their internal representation then the base is always 2. String representations of integers, now those have a base.
The following converts from decimal to octal, import java.util.Scanner; public class test { public static ...
I had this little homework assignment and I needed to convert decimal to octal and then octal to decimal. I did the first part and could not figure out the second to save my life. The first part went
An octal numeral consists of an ASCII digit 0 followed by one or more of the ASCII digits 0 through 7 interspersed with underscores, and can represent a positive, zero, or negative integer. As you can see, a bare 0 is considered as decimal. Whereas any (non-empty) sequence of digits preceded by 0 is considered as octal.
I am trying to use .format() in python I wish to print 1 to N with space padding so that all fields take the same width as the binary value. Below is what i have tried till now n=int(input())
int hexa = 0X1EF; // Octal declaration starts with 0 and possible chars are [0-7] int octal = 0757; // Binary representation starts with 0B or 0b and possible chars are [0-1] int binary = 0b111101111; If the number is string format then you can convert it into int using the below. String text = "0b111101111";
You could write your own, if you only care about three decimal places then set n to 3: def frac_to_oct(f, n=4): # store the number before the decimal point whole = int(f) rem = (f - whole) * 8 int_ = int(rem) rem = (rem - int_) * 8 octals = [str(int_)] count = 1 # loop until 8 * rem gives you a whole num or n times while rem and count < n: count += 1 int_ = int(rem) rem = (rem - int_) * 8 ...