exponent

问题 console.log(2E-12); // returns 2E-12 console.log(2E12); // returns 2000000000000 Why does line one return 2E-12 and not the same as line two. Is this an illegal way of using the exponent? 回答1: From the ECMAScript specification of toString applied to the Number type: 7. If 0 < n ≤ 21, return the String consisting of the most significant n digits of the decimal representation of s, followed by a decimal point ‘.’, followed by the remaining k−n digits of the decimal representation of s. 8. If −6

问题 I have the following string: "3.39112632978e+001" which I need to convert to float. WolframAlpha says that the result of this value is 33.9112632978 which evidently I should get somehow and I couldn't figure out how. Single.Parse("3.39112632978e+001") gives 3.39112624E+12 Double.Parse("3.39112632978e+001") gives 3391126329780.0 float.Parse("3.39112632978e+001") gives 3.39112624E+12 What should I do? 回答1: You are experiencing a localization issue wherein the . is being interpreted as a

问题 How do you implement exponents in JQuery? I'm modifying this: $.fn.sumValues = function() { var sum = 0; this.each(function() { if ( $(this).is(':input') ) { var val = $(this).val(); } else { var val = $(this).text(); } sum += parseFloat( ('0' + val).replace(/[^0-9-\.]/g, ''), 10 ); }); return sum; and $(document).ready(function() { $('input.price').bind('keyup', function() { $('span.total').html( $('input.price').sumValues() ); }); to calculate a more complex formula of summations from user

问题 I'm going to rely on the saying that no question is a dumb question, but I have a slightly dumb one to ask. EDIT: Seems that this question has been asked and answered a few times on here already, though using titles I didn't come across when searching for duplicates. Here are some related posts: Double to string conversion without scientific notation How to convert double to string without the power to 10 representation (E-05) Place with an answer (Jon Skeet): http://www.yoda.arachsys.com

问题 This is my code, trying to convert the second field of the line from exponential into float. outputrrd = processrrd.communicate() (output, error) = outputrrd output_lines = output.split('\n') for line in output_lines: m = re.search(r"(.*): ", line) if m != None: felder = line.split(': ') epoch = felder[0].strip(':') utc = epoch2normal(epoch).strip("\n") #print felder[1] data = float(felder[1]) float_data = data * 10000000 print float_data resultslist.append( utc + ' ' + hostname + ' ' + float

问题 This question already has answers here : Closed 9 years ago . Possible Duplicate: The most efficient way to implement an integer based power function pow(int, int) How can I calculate powers with better runtime? E.g. 2^13. I remember seeing somewhere that it has something to do with the following calculation: 2^13 = 2^8 * 2^4 * 2^1 But I can't see how calculating each component of the right side of the equation and then multiplying them would help me. Any ideas? Edit: I did mean with any base

问题 This question already has answers here : Closed 7 years ago . Possible Duplicate: How to check if a number is a power of 2 How could I write a method that would return true if passed in the value 2, 4, 8, 32, 64, and so on? 回答1: This is probably the best way: ((value & -value) == value) 回答2: Might want to look at this if you need a fast algorithm: http://en.wikipedia.org/wiki/Power_of_two#Fast_algorithm_to_check_if_a_positive_number_is_a_power_of_two 来源： https://stackoverflow.com/questions

问题 Given two integers: a <- 1L b <- 1L As I would expect, adding, subtracting, or multiplying them also gives an integer: class(a + b) # [1] "integer" class(a - b) # [1] "integer" class(a * b) # [1] "integer" But dividing them gives a numeric: class(a / b) # [1] "numeric" I think I can understand why: because other combinations of integers (e.g. a <- 2L and b <- 3L ) would return a numeric, it is the more general thing to do to always return a numeric. Now onto exponentiation: class(a ^ b) # [1]

问题 The goal is to calculate large digit numbers raised to other large digit numbers, e.g., 100 digit number raised to another 100 digit number, using recursion. My plan was to recursively calculate exp/2, where exp is the exponent, and making an additional calculation depending on if exp is even or odd. My current code is: def power(y, x, n): #Base Case if x == 0: return 1 #If d is even if (x%2==0): m = power(y, x//2, n) #Print statment only used as check print(x, m) return m*m #If d is odd else

问题 public static int exponent(int baseNum) { int temp = baseNum *= baseNum; return temp * exponent(baseNum); } Right now the method above does n * n into infinity if I debug it, so it still works but I need this recursive method to stop after 10 times because my instructor requires us to find the exponent given a power of 10. The method must have only one parameter, here's some examples of calling exponent: System.out.println ("The power of 10 in " + n + " is " + exponent(n)); So output should