integer

问题 I'm using phpMyAdmin to create my table structures. I can read from the documentation pages on MySQL about size limits for Integer Types: MySQL Integer Types Reference So here is where I'm getting a little confused with creating a column. I want to create a column in the table: tbl_note_categories called notescounter I don't foresee myself creating thousands of noteids in the tbl_notes with any specific categoryid . But I do believe I'd create hundreds of notes to each categoryid. I'm at that

问题 I'm under the assumption that INT(1) is the exact same thing as TINYINT(1) but I really have no idea. Whenever I've had values that can only be a single integer (e.g. a value 0-9), I've always just used INT(1) to say it's an integer and it will only be one character, which I assume means that it could only be a value 0 through 9 (please explain this to me if I'm wrong). I've always just ignored the other types of INT that you can cast the number as. I'm no MySQL savvy and tend to avoid the

问题 I have an array with integers of values from 0 to 100. I wish to remove integers that are less than number X and keep the ones that are equal or greater than number X. 回答1: A little ugly using the clunky create_function , but straight forward: $filtered = array_filter($array, create_function('$x', 'return $x >= $y;')); For PHP >= 5.3: $filtered = array_filter($array, function ($x) { return $x >= $y; }); Set $y to whatever you want. 回答2: Smarter than generating an array that is too big then

问题 Trying to figure out how to write a recursive predicate divide_by(X, D, I, R) that takes as input a positive integer X and a divisor D, and returns the answer as the whole number part I and the remainder part R, however, I can't seem to get my head around Prolog. How would I go about doing this? 回答1: There are predefined evaluable functors for this. (div)/2 and (mod)/2 always rounding down. Recommended by LIA-1, Knuth etc. (//)/2 and (rem)/2 rounding toward zero (actually, it's implementation

问题 My understanding of the Integer.TryParse() function was that it tried to parse an integer from the passed in string and if the parse failed the result integer would remain as it did before. I have an integer with a default value of -1 which I would like to remain at -1 if the parse fails. However the Integer.TryParse() function on failing to parse is changing this default value to zero. Dim defaultValue As Integer = -1 Dim parseSuccess As Boolean = Integer.TryParse("", defaultValue) Debug

问题 This question already has answers here : Determine if a C string is a valid int in C (5 answers) Closed 6 years ago . I have a program which accepts an integer from the user and uses this number in an addition operation. The code which I am using to accept the number is this: scanf("%d", &num); How can I validate the input such that if the user enters a letter or a number with the decimal point, an error message is displayed on the screen? 回答1: You should use scanf return value. From man

问题 Say I wanted to create an array (NOT list) of 1,000,000 twos in python, like this: array = [2, 2, 2, ...... , 2] What would be a fast but simple way of doing it? 回答1: Is this what you're after? # slower. twosArr = array.array('i', [2] * 1000000) # faster. twosArr = array.array('i', [2]) * 1000000 You can get just a list with this: twosList = [2] * 1000000 -- EDITED -- I updated this to reflect information in another answer. It would appear that you can increase the speed by a ratio of ~ 9 : 1

问题 What is the proper way to deal with leading zeros in Ruby? 0112.to_s => "74" 0112.to_i => 74 Why is it converting 0112 into 74 ? How can convert 0112 to a string "0112" ? I want to define a method that takes integer as a argument and returns it with its digits in descending order. But this does not seem to work for me when I have leading zeros: def descending_order(n) n.to_s.reverse.to_i end 回答1: A numeric literal that starts with 0 is an octal representation, except the literals that start

问题 I wish to change an integer such as 23457689 to 689, 12457245 to 245 etc. I do not require the numbers to be rounded and do not wish to have to convert to String. Any ideas how this can be done in Python 2.7? 回答1: Use the % operation: >>> x = 23457689 >>> x % 1000 689 % is the mod (i.e. modulo) operation. 回答2: To handle both positive and negative integers correctly: >>> x = -23457689 >>> print abs(x) % 1000 689 As a function where you can select the number of leading digits to keep: import

问题 Is there a function for integer exponentiation in OCaml? ** is only for floats. Although it seems to be mostly accurate, isn't there a possibility of precision errors, something like 2. ** 3. = 8. returning false sometimes? Is there a library function for integer exponentiation? I could write my own, but efficiency concerns come into that, and also I'd be surprised if there isn't such a function already. 回答1: Regarding the floating-point part of your question: OCaml calls the underlying