variable-assignment

问题 I have two classes, call them A and B. They both contain a Loader object. In class A I load content into the Loader object. public class A { var loader:Loader; public function A():void { loader = new Loader(); this.addChild(loader); loader.load(...); } } public class B() { var loader:Loader; public function B():void { loader = new Loader(); this.addChild(loader); } } I need to now assign A's Loader to B's Loader (after the load is complete). In some other class I have an instance of A and B.

问题 I am using multiple redirects, and I would like to redirect from A->B->C->A. So in B, I save path A as @previouspage = request.referer and so @previouspage = A at this point, but when I call @previouspage in C, it doesn't bring the hard value saved in B, but finds its own relative request.referer, which is B. So in C, @previouspage = B (because I think variables in Ruby are soft-links) How would I just save whatever the value of request.referer was at point B, and then save that URL into a

问题 I have a series of dictionaries that hold values that I want to combine at the end . The combination is messy so I can't just combine the dictionaries I have to work based on keys and make some careful choices. When I created the dictionaries I created them in the form item#Dict expecting to do something like for value in ['8','10','11','12','13','14','15','18','19','20']: dictName='item'+value+'Dict' but as I quickly learned dictName is a string not a reference to my dictionary. Is there a

问题 I checked the operator precedence in Python 3(https://docs.python.org/3/reference/expressions.html#operator-precedence), and found that there is no assignment(=). I want to know if assignment is an operator or not. If not, why there are so many "assignment operator" info when googled? What is the precedence relation with other real operators(bool operator, comparison operator, etc)? 回答1: No. An assignment is always a statement in Python. That's why things like assignment within if statements,

问题 I am trying to run an old python code made with python 2.7(?) in python 3.3 and I'm stuck on updating the code to run. It keeps telling me "'range' object does not support item assignment" and for the life of I cannot figure it out. The code is for a "50 states trivia" game I found on google. The error is at the line answer[i] = "%s " % flower[pick[i]].rstrip() pick = random.sample(range(50), 4) print("The state flower of %s is:" % state[pick[0]]) answer = range(4) for i in range(4): if i ==

问题 I get the number of a specific process with the help of this thread: How to count amount of processes with identical name currently running, using a batchfile I hope to assign the result of this command to a variable, then compare the variable with a number. My code is listed as below: @echo off setlocal enabledelayedexpansion set procName=chrome.exe set a=tasklist /FI "IMAGENAME eq %procName%" 2>NUL | find /I /C "%procName%" if !a! equ 1 ( echo !a! echo %procName% starts to run... ) else (

问题 I can't figure out how to tell C that I want a pointer that will not move. It will always point to the same array. That said, the array members are not constant, but the array itself is global and so, it is at a fixed position. So, when I code this: #include <stdio.h> int v[2]={0, 1}; const int *cpv=v; int main(void) { v[1]=2; printf("%d\n", v[1]); *(cpv+1)=3; printf("%d\n", v[1]); cpv[1]=4; printf("%d\n", v[1]); } And get this errors: constp.c: In function ‘main’: constp.c:9: error:

问题 Assignment: >>> a = ['spam'] >>> b = ['spam'] >>> a is b False Reference: >>> c = ['spam'] >>> d = c >>> c is d True What is the difference between above both? Why assignment results False ? Why reference results True ? 回答1: Your first snippet creates two unique list objects, which are not the same. Hence a is b returns false because a and b are pointing to distinct objects: +------+ a ------> | list | +------+ +------+ b ------> | list | +------+ Your second snippet creates a single list

问题 Assignment: >>> a = ['spam'] >>> b = ['spam'] >>> a is b False Reference: >>> c = ['spam'] >>> d = c >>> c is d True What is the difference between above both? Why assignment results False ? Why reference results True ? 回答1: Your first snippet creates two unique list objects, which are not the same. Hence a is b returns false because a and b are pointing to distinct objects: +------+ a ------> | list | +------+ +------+ b ------> | list | +------+ Your second snippet creates a single list

问题 suppose I have a data object that is uploaded by the user: data <- reactive({ inFile <- input$file if (is.null(inFile)) return(NULL) data <- read.csv(inFile$datapath) return(data) }) And suppose I want to delete columns in the dataset. I want to set it to global assignment so that I can run the UI multiple times and have each effect saved in the object. dataset <- reactive({ file1 <- data() file1[,input$deletecols] <<- NULL return(file1) }} }) However, when I run this, I get the error: