long-integer

问题 So this the line with the precision error fault; A[i]= m % 3; m is long A is int[]; And my error is error: possible loss of precision A[i]= m % 3. required int found long. How can I have error when the only potential answers are 0,1,2? Isn't there another way than declaring A as long[]? It's a potentially big array so I don't want that (in fact I would even prefer for A to be short[]) Also I tried error: A[i]= m % 3L , but same result. 回答1: The compiler doesn't look at the result, it looks at

问题 I gather that to represent a 64-bit integer in R I need to use a double. That's fine but I need to read such an integer from a binary file where it is stored as Big Endian 64-bit two's-complement (a java long). I can of course read the two signed integers in 4 byte chunks like so a = readBin(file, integer(), size=4, endian="big") b = readBin(file, integer(), size=4, endian="big") But how do I combine them in R to get the double I require? 回答1: It is definitely better to read it in as two

问题 I have been testing out with this snippet of code: public class Test { public static int longToInt(long num) { return (int) (num); } public static void main(String[] args) { System.out.println(longToInt(100000000L)); } } I ran it but longToInt only returned 0. What is going on? 回答1: Casting a long to an int is done by removing the top 32 bits of the long . If the long value is larger than Integer.MAX_VALUE (2147483647) or smaller than Integer.MIN_VALUE (-2147483648), the net effect is that

问题 I have a Long array with these numbers: long[] = {1,2,3,5,6,7}; Notice that 4 is missing. What's the best way to test this array if any such gaps exist or not? 回答1: If you're guaranteed that arrays is ordered without any duplicate then you could check that in O(1) I think this code should work in this specific case :) //assume that given array is ordered and has no duplicated value long[] myarray = {5,6,7}; //no gap long[] myarray1 = {1,2,4}; //has gap long[] myarray2 = {10,11,12,13,14,15}; /

问题 Hey everyone this may turn out to be a simple stupid question, but one that has been giving me headaches for a while now. I'm reading data from a Named Binary Tag file, and the code is working except when I try to read big-endian numbers. The code that gets an integer looks like this: long NBTTypes::getInteger(istream &in, int num_bytes, bool isBigEndian) { long result = 0; char buff[8]; //get bytes readData(in, buff, num_bytes, isBigEndian); //convert to integer cout <<"Converting bytes to

问题 I need to compare an integer in a loop to the quotient of a long and a long . in order to not do integer division, if I understand correctly do I need to convert one of the longs into a double? long prime = primes[d]; int i = 1; // "inputNumber / prime" should not be integer division, while it is now. // How do I do this yet still compare it to an "int" afterwards? while (i < (inputNumber / prime)) { primes[i*prime] = 0; i++; } That's the code snippet. primes is an array filled with long s.

问题 Why is this program not working? I convert a byte array to long. Then from the long I convert back to a byte array. The resulting byte array is not the same as original. class Program { static void Main(string[] args) { byte[] myBytes = { 0, 0, 0, 32, 56, 99, 87, 34, 56, 56, 34, 33, 67 , 56, 66, 72, 1, 0, 0, 56, 0, 22}; long data = BitConverter.ToInt64(myBytes, 0); byte[] byteData = BitConverter.GetBytes(data); Console.WriteLine("byte array: " + BitConverter.ToString(myBytes)); Console

问题 I have this C program codes are completely and it's works very well, but not the correct result. I have them code source here in the below. Can anyone help me what is going on or what did I missing? Everyone are welcome to correct this English or code. #include <stdio.h> #include <stdlib.h> #include <math.h> int main (int argc, char *argv[]) { int n; int result; if (argc < 2) { printf ("Usage: p4 <number>\n"); } n = atoi (argv[1]); if (n < 2) { printf ("input number should be > 1\n"); }

问题 String a = "576055795"; long b = 10*Integer.parseInt(a); long c = 10*Long.parseLong(a); System.out.println(b); //Prints 1465590654 System.out.println(c); // Prints 5760557950 Why are they different? 回答1: Integer.parseInt() returns an int , which is a signed 32-bit integer. 10 is also an int ; multiplying 576055795 by 10 as ints overflows and yields an int , which is then promoted to a long . Long.parseLong() returns a long , which is a signed 64-bit integer. Multiplying it by 10 yields a long

问题 I've been trying to print out the Binary representation of a long long integer using C Programming My code is #include<stdio.h> #include <stdlib.h> #include<limits.h> int main() { long long number, binaryRepresentation = 0, baseOfOne = 1, remainder; scanf("%lld", &number); while(number > 0) { remainder = number % 2; binaryRepresentation = binaryRepresentation + remainder * baseOfOne; baseOfOne *= 10; number = number / 2; } printf("%lld\n", binaryRepresentation); } The above code works fine