increment

问题 I notice that a pre-increment/decrement operator can be applied on a variable (like ++count ). It compiles, but it does not actually change the value of the variable! What is the behavior of the pre-increment/decrement operators (++/--) in Python? Why does Python deviate from the behavior of these operators seen in C/C++? 回答1: ++ is not an operator. It is two + operators. The + operator is the identity operator, which does nothing. (Clarification: the + and - unary operators only work on

问题 Consider the following numeric widget in an R Shiny app: numericInput("val", "Enter value:", value = 50, min = 0, step = 5) If you click on the up/down arrows in the widget when the app is run, the value will increase or decrease by 5 (0, 5, 10, 15,...) as expected. Now consider changing the min value to 1: numericInput("val", "Enter value:", value = 50, min = 1, step = 5) If you now click on the up/down arrows, the value will still increase/decrease by 5, but start from 1, creating the

问题 This question already has answers here : Closed 8 years ago . So I am stuck with this question. int i=5,a; a=++i + i++ + ++i + i++ - --i; printf("%d",a); According to me 'a' should be 20. a=6+6+8-8 However, on execution I found the answer to be 18. What I am doing wrong? A step by step explanation would be helpful. 回答1: This is undefined behavior . A variable cannot be changed more than once between sequence points. Your program can output anything. Any answer stating otherwise is wrong. 回答2:

问题 This question already has answers here : Why are these constructs using pre and post-increment undefined behavior? (14 answers) Closed 5 years ago . I don't know if this is compiler specific but when I tried running the two expressions in DevC++ When i=c=b=0; i=i++ + ++c gives 2 whereas i=++i + c++ gives 1 But b=i++ + ++c and b=++i + ++c produces the result 1 for both expressions. I do know that incrementing a variable twice in the same expression results in an undefined value according to

问题 Is it possible to increment a variable or array by 1 each time the script is executed. An example would be, I execute the script and it sets a variable to $increment = '1'; I then close the script and come back 30 min later to execute the script again. This time the variable needs to be $increment = '2'; The script will be run from cron every 30 minutes for a total of 8 times. After the 8th execution the variable would need to be rewound back to $increment = '1'; Is this possible through

问题 I've looked online for hours trying to see if I could find a solution and while I have found many solutions my instructions from my professor are as follows: Write a program to estimate PI (π) using the following series. This problem is also described in the text as problem 5.25 at the end of chapter 5. If you are unfamiliar with series, problem 5.24 is a single pass through a series and the solution is posted in the Homework 3 course module. π=4*(1-1/3+1/5-1/7+1/9-1/11+⋯〖-1〗^(i+1)/(2i-1))

问题 I want to write a for loop that will iterate over the powers of 2 for each loop. For example I would want a range like this: 2, 4, 8, 16, ... , 1024 How could I do this? 回答1: counter = 2 while counter <= 1024: print counter counter *= 2 回答2: You'll need to create your own function: def doubling_range(start, stop): while start < stop: yield start start <<= 1 This uses a left-shift operation; you could also use start *= 2 if you find that clearer. Demo: >>> def doubling_range(start, stop): ...

问题 This question already has answers here : How do the post increment (i++) and pre increment (++i) operators work in Java? (14 answers) Closed 2 years ago . Day after tomorrow is my exam for Computers (JAVA) and I have a big problem in the above title. I understood what does post and pre increment and decrement means. But I can no understand what to do when the matter comes to a complex, long statement. One example for such question is below. class java_1 { public void main() { int x = 4; x +=

问题 public class Sample { public static void main(String[] args) throws Exception { //part 1 int i=1; i=i++; i=++i; i=i++; System.out.println(i); //part 2 i=1; int a=i++; a=++i; a=i++; System.out.println(a+"\n"+i); } } Output 2 3 4 Yesterday my friend asked this question. I little bit confused about this. part 1 prints the i value as 2 . Post increment is not working here. But in the part 2 , it works. I can understand the part 2 but i have confused in part 1 . How actually it works? Can anybody

问题 This question already has answers here : Why are these constructs using pre and post-increment undefined behavior? (14 answers) printf(“%d %d %d\n”,++a, a++,a) output [duplicate] (3 answers) Closed 2 years ago . If n has the value 5 then output: printf("%d %d", n++, ++n); //should be 5 and 7 But I get as output 6 and 7. 回答1: Multiple unsequenced modifications result to such kind of Undefined Behavior . There are tons of results if you search for it, e.g. this question. Next time compile with