operators

问题 import java.lang.Math; import java.math.BigInteger; import java.math.BigDecimal; public class Main { public static void main(String[] args) { int e1 = 20, d = 13; BigInteger C = BigDecimal.valueOf(e1).toBigInteger(); BigInteger po = C.pow(d); System.out.println("pow is:" + po); int num = 11; BigInteger x = po; BigInteger n = BigDecimal.valueOf(num).toBigInteger(); BigInteger p, q, m; System.out.println("x: " + x); q=(x / n); p=(q * n); m=(x - p); System.out.println("mod is:" + m); } } I've

问题 When checking if a integer is the same or above a current number.. so I type if (5 => 6) { //Bla } but it shows this as a error. Why? Isn't it exactly the same as if (5 >= 6) { //Bla } 回答1: The reason why it does not work is because => is not equivalent to >= . => is used in a lambda expression. Like : (int x, string s) => s.Length > x I do agree it is annoying. Before lambda expressions I used to get it wrong sometimes. Now I always know that one ( => ) is a lambda expression and and other (

问题 This question already has answers here : What are the rules for evaluation order in Java? (5 answers) If parenthesis has a higher precedence then why is increment operator solved first? (5 answers) Closed 5 years ago . In this code: int y = 10; int z = (++y * (y++ + 5)); What I expected First y++ + 5 will be executed because of the precedence of the innermost parentheses. So value of y will be 11 and the value of this expression will be 15. Then ++y * () will be executed. So 12 * 15 = 180. So

问题 This question already has answers here : What are the rules for evaluation order in Java? (5 answers) If parenthesis has a higher precedence then why is increment operator solved first? (5 answers) Closed 5 years ago . In this code: int y = 10; int z = (++y * (y++ + 5)); What I expected First y++ + 5 will be executed because of the precedence of the innermost parentheses. So value of y will be 11 and the value of this expression will be 15. Then ++y * () will be executed. So 12 * 15 = 180. So

问题 This question already has answers here : What are the rules for evaluation order in Java? (5 answers) If parenthesis has a higher precedence then why is increment operator solved first? (5 answers) Closed 5 years ago . In this code: int y = 10; int z = (++y * (y++ + 5)); What I expected First y++ + 5 will be executed because of the precedence of the innermost parentheses. So value of y will be 11 and the value of this expression will be 15. Then ++y * () will be executed. So 12 * 15 = 180. So

问题 I have the following code: unsigned char x = 255; printf("%x\n", x); // ff unsigned char tmp = x << 7; unsigned char y = tmp >> 7; printf("%x\n", y); // 1 unsigned char z = (x << 7) >> 7; printf("%x\n", z); // ff I would have expected y and z to be the same. But they differ depending on whether a intermediary variable is used. It would be interesting to know why this is the case. 回答1: This little test is actually more subtle than it looks as the behavior is implementation defined: unsigned

问题 I can't understand output of this program: #include<iostream> using namespace std; int main() { int x = 1 , y = 1, z = 1; cout << ( ++x || ++y && ++z ) << endl; //outputs 1; cout << x << " " << y << " " << z ; //x = 2 , y = 1 , z = 1; return 0; } Output: 1 2 1 1 If || is evaluated first then this output is fine, however this article says that && has a higher precedence than || , and thus it must be evaluated first. If this is the case then according to me output should be: 1 1 2 2 as ++y && +

问题 I can't understand output of this program: #include<iostream> using namespace std; int main() { int x = 1 , y = 1, z = 1; cout << ( ++x || ++y && ++z ) << endl; //outputs 1; cout << x << " " << y << " " << z ; //x = 2 , y = 1 , z = 1; return 0; } Output: 1 2 1 1 If || is evaluated first then this output is fine, however this article says that && has a higher precedence than || , and thus it must be evaluated first. If this is the case then according to me output should be: 1 1 2 2 as ++y && +

问题 We know that double-colon ( :: ) is used to get function (callable) reference in Kotlin, e.g. String::compareTo , "string"::compareTo . In Java we use SomeClass.class and someInstance.getClass() to get the class. Why in Kotlin we use SomeClass::class and someInstance::class while class is not a function/method? println(String::compareTo) // output: fun kotlin.String.compareTo(kotlin.String): kotlin.Int println("string".compareTo("strong")) // output: -6 println(String::class) // output: class

问题 Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 3 years ago . In my code, I'm trying to make a not less than operator. I can't do a !< b ... what can I do? Is there any package / method I can use? 回答1: The opposite of a<b is a>=b , not a>b . EDIT: The syntax you were looking for is !(a<b) 回答2: Oh, I am so stupid! I could just do a >= b ! 来源：