scope

问题 This question originates from my previous question Why do fully qualified names have to be used when dynamically assessing namespaced elements? and another question Use php namespace inside function. Both of these questions boil down to the following explanation by php.net: Importing is performed at compile-time, and so does not affect dynamic class, function or constant names.... ....The use keyword must be declared in the outermost scope of a file (the global scope) or inside namespace

问题 I am struggling here. i cannot figure out how to make the button functions read the $listbox1.selectedItems and $listbox2.selectedItem whats wrong here? if i try to look at the $listbox items after the OK button been pressed it will show me but not if i call the function button, ive deleted some unneccesery code parts. $form = New-Object System.Windows.Forms.Form $form.StartPosition = 'CenterScreen' $button1 = New-Object System.Windows.Forms.Button $button1.Text = 'Link' $button2 = New-Object

问题 So let's say I have a lua file, and at the top, I define a variable outside of any function, but I call it local local x = 1 Is there any difference between that local x, and a global x? 回答1: Yes, as it is local to the chunk that it is created in. Lua handles a chunk as the body of an anonymous function with a variable number of arguments (see §3.4.11). As such, chunks can define local variables, receive arguments, and return values. Moreover, such anonymous function is compiled as in the

问题 Closed . This question is opinion-based. It is not currently accepting answers. Want to improve this question? Update the question so it can be answered with facts and citations by editing this post. Closed 10 days ago . Improve this question I am teaching C programming and I am quite strict regarding variable declaration. As we use the C11 standard and knowing that the students would later use other programming language, I teach this rule: Any declared symbol should have its scope as narrow

问题 This question already has answers here : Can a local variable's memory be accessed outside its scope? (20 answers) Closed 5 years ago . There's a PDF I'm reading which says that a pointer is invalid after it passes out of scope. See slide #14 in the file below: http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-087-practical-programming-in-c-january-iap-2010/lecture-notes/MIT6_087IAP10_lec05.pdf Now I wrote almost the exact same code below in C++ with Dev-C++ compiler:

问题 I have the following functions in Python3.7 def output_report(): sheet_dict = {1: 'All', 2: 'Wind', 3: 'Soalr'} sheet_name = sheet_dict[sheet_num] file_name = f'{file_name}_{sheet_name}_row_{row_num}.csv' if row_num else f'{file_name}_{sheet_name}.csv' return file_name def test_file(x): file_name = output_report(sheet_num) return f'{x}_{file_name}' def good_func(): sheet_num = 2 row_num = 2 a = test_file('new file') return a when I call: good_func() It raises an error that: NameError: name

问题 I have a data frame that I initialize out of scope of a local method. I would like to do as follows: def outer_method(): ... do outer scope stuff here df = pd.DataFrame(columns=['A','B','C','D']) def recursive_method(arg): ... do local stuff here # func returns a data frame to be appended to empty data frame results_df = func(args) df.append(results_df, ignore_index=True) return results recursive_method(arg) return df However, this does NOT work. The df is always empty if I append to it this

问题 General problem I've noticed several times that push-pop scopes that have already been popped appear to affect the time that a check-sat in a subsequent scope needs. That is, assume a program with multiple (potentially arbitrarily nested) push-pop scopes, each of which contain a check-sat command. Furthermore, assume that the second check-sat takes 10s, whereas the first one takes only 0.1s. ... (push) (assert (not P)) (check-sat) ; Could be sat, unsat or unknown (pop) ... (push) (assert (not

问题 I'm a bit confused about how scope & variable assignment within a function seems to change when assigning a default parameter value for that function. For example, when this function has a default value assigned to parameter i , the output array variable appears to be block-scoped when inspected with Chrome Dev Console,: function steps(n, i = 40) { var output = [n]; } steps(10, 20); However, by removing the default parameter value for i , the output array variable is scoped locally: function

问题 function Job(name, cronString, task) { "use strict"; this.name = name; this.cronString = cronString; this.isReady = false; this.task = task; } Job.prototype.performTask = (db, winston) => { "use strict"; const Promise = require("bluebird"); let that = this; return new Promise((resolve, reject) => { let output = ""; let success = true; try { output = that.task(); } catch(error) { success = false; reject(error); } if(success) { resolve(output); } }); }; module.exports = Job; Javascript newbie