local-variables

Issue with setting value for Python class inherited from multiprocessing.Process

别来无恙 提交于 2019-12-06 10:50:23
问题 Why this code import multiprocessing import time class Bot(multiprocessing.Process): def __init__(self): self.val = 0 multiprocessing.Process.__init__(self) def setVal(self): self.val = 99 def run(self): while True: print 'IN: ', self.val time.sleep(2) if __name__ == '__main__': bot = Bot() bot.start() bot.setVal() while True: print 'OUT: ', bot.val time.sleep(2) gives following output? OUT: 99 IN: 0 OUT: 99 IN: 0 OUT: 99 IN: 0 OUT: 99 IN: 0 OUT: 99 IN: 0 OUT: 99 IN: 0 ... As you may guess i

Use var in field declaration

試著忘記壹切 提交于 2019-12-06 08:38:29
So starting in Java 9, we can use var to declare local variables: var s = "cool"; is there a way to use a similar construct when declaring fields? class Container { final var s = "cool"; // does not compile tmk } doesn't seem like it from what I can tell. Jacob G. is there a way to use a similar construct when declaring fields? No. According to JEP 286: Local-Variable Type Inference : This treatment would be restricted to local variables with initializers, indexes in the enhanced for-loop, and locals declared in a traditional for-loop; it would not be available for method formals, constructor

Parameter scope vs local variable scope?

左心房为你撑大大i 提交于 2019-12-06 07:12:12
问题 I was reading my AP cs book and it talked about three types of variables: •Instance variables •Local variables •Parameters Instance variables are visible throughout the class etc... Parameters are only usable within the method and so are local variables . . . Therefore my question is why would they classify Parameters and Local variables as different categories of variables if they contain the same scope. . . Despite the different uses of them. 回答1: Because they don't necessarily have the

Getting local variables

一个人想着一个人 提交于 2019-12-05 12:53:20
When getting a stacktrace as error report from an application that is already deployed, it would be helpful to also get the actual variable values to reconstruct the system's state at the point before the exception was thrown. Is anything like that feasible in Java and how could one do that? Cheers, Max I'm pretty sure you cannot get the local variables in the stacktrace as the output is built from instance of StackTraceElement which only contains, the class, the file, the method and the line number (see http://download.oracle.com/javase/6/docs/api/java/lang/StackTraceElement.html ). Have a

Local variables on stack

杀马特。学长 韩版系。学妹 提交于 2019-12-05 05:25:59
问题 To understand the stack frame concept, I wrote a little program for my own. First I will show you the code, a little sketch about it and then I will present my question: So, the program: int check_pw(char *password){ int valid = 0; char buffer[10]; strcpy(buffer, password); if(strcmp(buffer, "a") == 0){ valid = 1; } return valid; } int main(int argc, char *argv[]){ if(check_pw(argv[1])){ printf("OK\n"); } else{ printf("Wrong password\n"); } } I give the password as a command-line argument.

How do I value-initialize a Type* pointer using Type()-like syntax?

假装没事ソ 提交于 2019-12-04 23:38:17
Variables of built-in types can be value-initialized like this : int var = int(); this way I get the default value of int without hardcoding the zero in my code. However if I try to do similar stuff for a pointer: int* ptr = int*(); the compiler (Visual C++ 10) refuses to compile that (says type int unexpected ). How do I value-initialize a pointer in similar manner? How do I value-initialize a Type* pointer using Type()-like syntax? You cannot. The syntax T() is defined in 5.2.3/1,2 (C++03, slightly different wording in C++11 FDIS). In particular the second paragraph states: The expression T(

How to list local-variables in Ruby?

末鹿安然 提交于 2019-12-04 22:17:56
def method a = 3 b = 4 some_method_that_gives # [a, b] end Nakilon local_variables It outputs array of symbols, presenting variables. In your case: [:a, :b] local_variables lists local variables but it lists them before they are defined. See this: p local_variables a = 1 p local_variables this outputs [:a] [:a] which may not be what you expect. Contrast with defined? p defined? a a = 1 p defined? a which outputs the more anticipated nil "local-variable" 来源: https://stackoverflow.com/questions/4487326/how-to-list-local-variables-in-ruby

Lambda assigning local variables

空扰寡人 提交于 2019-12-04 17:42:30
问题 Consider the following source: static void Main(string[] args) { bool test; Action lambda = () => { test = true; }; lambda(); if (test) Console.WriteLine("Ok."); } It should compile, right? Well, it doesn't. My question is: according to C# standard, should this code compile or is this a compiler bug? The error message: Use of unassigned local variable 'test' Note: I know , how to fix the error and i partially know , why does it happen. However, the local variable is assigned unconditionally

Javascript - local scope objects not accessible from nested function

大兔子大兔子 提交于 2019-12-04 17:28:19
I am trying to have a function grab an object from a php file on another page. I'm using the jQuery ajax function to to do the json grab, which is working correctly. The issue is when I try to return that object from the function. The first time I log the object (from within the success function) it is correct in the console, but the returned object from the function getGantt() logs as "undefined". How do I get this object out of the function? My code: function getGantt(requestNumber){ var ganttObject; $.ajax({ type: "POST", url: "get_gantt.php", data: {request_number: requestNumber}, success:

Can static local variables cut down on memory allocation time?

柔情痞子 提交于 2019-12-04 16:45:31
问题 Suppose I have a function in a single threaded program that looks like this void f(some arguments){ char buffer[32]; some operations on buffer; } and f appears inside some loop that gets called often, so I'd like to make it as fast as possible. It looks to me like the buffer needs to get allocated every time f is called, but if I declare it to be static, this wouldn't happen. Is that correct reasoning? Is that a free speed up? And just because of that fact (that it's an easy speed up), does