My University is testing blackboard ultra. I generate the exercises with
exams2blackboard("ProbEjemploPrueba9", n=10, fix_pre = TRUE, converter = "pandoc", encoding = "utf8", eval = list(partial = T, negative = T, rule = "false"))
but when I use the $ in $A$, it does not leave space, it appears like this;
"IfA is the event .... "instead of "If A is the event.... "
I attach the image.
The solution is to remove $, because in this case nothing happens, but maybe other times it is necessary.
Would there be any solution to fix that typo?
Related
I working on a language similar to ruby called gaiman and I'm using PEG.js to generate the parser.
Do you know if there is a way to implement heredocs with proper indentation?
xxx = <<<END
hello
world
END
the output should be:
"hello
world"
I need this because this code doesn't look very nice:
def foo(arg) {
if arg == "here" then
return <<<END
xxx
xxx
END
end
end
this is a function where the user wants to return:
"xxx
xxx"
I would prefer the code to look like this:
def foo(arg) {
if arg == "here" then
return <<<END
xxx
xxx
END
end
end
If I trim all the lines user will not be able to use a string with leading spaces when he wants. Does anyone know if PEG.js allows this?
I don't have any code yet for heredocs, just want to be sure if something that I want is possible.
EDIT:
So I've tried to implement heredocs and the problem is that PEG doesn't allow back-references.
heredoc = "<<<" marker:[\w]+ "\n" text:[\s\S]+ marker {
return text.join('');
}
It says that the marker is not defined. As for trimming I think I can use location() function
I don't think that's a reasonable expectation for a parser generator; few if any would be equal to the challenge.
For a start, recognising the here-string syntax is inherently context-sensitive, since the end-delimiter must be a precise copy of the delimiter provided after the <<< token. So you would need a custom lexical analyser, and that means that you need a parser generator which allows you to use a custom lexical analyser. (So a parser generator which assumes you want a scannerless parser might not be the optimal choice.)
Recognising the end of the here-string token shouldn't be too difficult, although you can't do it with a single regular expression. My approach would be to use a custom scanning function which breaks the here-string into a series of lines, concatenating them as it goes until it reaches a line containing only the end-delimiter.
Once you've recognised the text of the literal, all you need to normalise the spaces in the way you want is the column number at which the <<< starts. With that, you can trim each line in the string literal. So you only need a lexical scanner which accurately reports token position. Trimming wouldn't normally be done inside the generated lexical scanner; rather, it would be the associated semantic action. (Equally, it could be a semantic action in the grammar. But it's always going to be code that you write.)
When you trim the literal, you'll need to deal with the cases in which it is impossible, because the user has not respected the indentation requirement. And you'll need to do something with tab characters; getting those right probably means that you'll want a lexical scanner which computes visible column positions rather than character offsets.
I don't know if peg.js corresponds with those requirements, since I don't use it. (I did look at the documentation, and failed to see any indication as to how you might incorporate a custom scanner function. But that doesn't mean there isn't a way to do it.) I hope that the discussion above at least lets you check the detailed documentation for the parser generator you want to use, and otherwise find a different parser generator which will work for you in this use case.
Here is the implementation of heredocs in Peggy successor to PEG.js that is not maintained anymore. This code was based on the GitHub issue.
heredoc = "<<<" begin:marker "\n" text:($any_char+ "\n")+ _ end:marker (
&{ return begin === end; }
/ '' { error(`Expected matched marker "${begin}", but marker "${end}" was found`); }
) {
const loc = location();
const min = loc.start.column - 1;
const re = new RegExp(`\\s{${min}}`);
return text.map(line => {
return line[0].replace(re, '');
}).join('\n');
}
any_char = (!"\n" .)
marker_char = (!" " !"\n" .)
marker "Marker" = $marker_char+
_ "whitespace"
= [ \t\n\r]* { return []; }
EDIT: above didn't work with another piece of code after heredoc, here is better grammar:
{ let heredoc_begin = null; }
heredoc = "<<<" beginMarker "\n" text:content endMarker {
const loc = location();
const min = loc.start.column - 1;
const re = new RegExp(`^\\s{${min}}`, 'mg');
return {
type: 'Literal',
value: text.replace(re, '')
};
}
__ = (!"\n" !" " .)
marker 'Marker' = $__+
beginMarker = m:marker { heredoc_begin = m; }
endMarker = "\n" " "* end:marker &{ return heredoc_begin === end; }
content = $(!endMarker .)*
I have stumbled upon this line of code and I am not sure what the [ ? ] part represents (my guess is it's a sort of a wildcard but I searched it for a while and couldn't find anything):
['?'] = function() return is_canadian and "eh" or "" end
I understand that RHS is a functional ternary operator. I am curious about the LHS and what it actually is.
Edit: reference (2nd example):
http://lua-users.org/wiki/SwitchStatement
Actually, it is quite simple.
local t = {
a = "aah",
b = "bee",
c = "see",
It maps each letter to a sound pronunciation. Here, a need to be pronounced aah and b need to be pronounced bee and so on. Some letters have a different pronunciation if in american english or canadian english. So not every letter can be mapped to a single sound.
z = function() return is_canadian and "zed" or "zee" end,
['?'] = function() return is_canadian and "eh" or "" end
In the mapping, the letter z and the letter ? have a different prononciation in american english or canadian english. When the program will try to get the prononciation of '?', it will calls a function to check whether the user want to use canadian english or another english and the function will returns either zed or zee.
Finally, the 2 following notations have the same meaning:
local t1 = {
a = "aah",
b = "bee",
["?"] = "bee"
}
local t2 = {
["a"] = "aah",
["b"] = "bee",
["?"] = "bee"
}
If you look closely at the code linked in the question, you'll see that this line is part of a table constructor (the part inside {}). It is not a full statement on its own. As mentioned in the comments, it would be a syntax error outside of a table constructor. ['?'] is simply a string key.
The other posts alreay explained what that code does, so let me explain why it needs to be written that way.
['?'] = function() return is_canadian and "eh" or "" end is embedded in {}
It is part of a table constructor and assigns a function value to the string key '?'
local tbl = {a = 1} is syntactic sugar for local tbl = {['a'] = 1} or
local tbl = {}
tbl['a'] = 1
String keys that allow that convenient syntax must follow Lua's lexical conventions and hence may only contain letters, digits and underscore. They must not start with a digit.
So local a = {? = 1} is not possible. It will cause a syntax error unexpected symbol near '?' Therefor you have to explicitly provide a string value in square brackets as in local a = {['?'] = 1}
they gave each table element its own line
local a = {
1,
2,
3
}
This greatly improves readability for long table elements or very long tables and allows you maintain a maximum line length.
You'll agree that
local tbl = {
z = function() return is_canadian and "zed" or "zee" end,
['?'] = function() return is_canadian and "eh" or "" end
}
looks a lot cleaner than
local tbl = {z = function() return is_canadian and "zed" or "zee" end,['?'] = function() return is_canadian and "eh" or "" end}
as a followup question to lua tables - allowed values and syntax:
I need a table that equates large numbers to strings. The catch seems to be that strings with punctuation are not allowed:
local Names = {
[7022003001] = fulsom jct, OH
[7022003002] = kennedy center, NY
}
but neither are quotes:
local Names = {
[7022003001] = "fulsom jct, OH"
[7022003002] = "kennedy center, NY"
}
I have even tried without any spaces:
local Names = {
[7022003001] = fulsomjctOH
[7022003002] = kennedycenterNY
}
When this module is loaded, wireshark complains "}" is expected to close "{" at line . How can I implement a table with a string that contains spaces and punctuation?
As per Lua Reference Manual - 3.1 - Lexical Conventions:
A short literal string can be delimited by matching single or double quotes, and can contain the (...) C-like escape sequences (...).
That means the short literal string in Lua is:
local foo = "I'm a string literal"
This matches your second example. The reason why it fails is because it lacks a separator between table members:
local Names = {
[7022003001] = "fulsom jct, OH",
[7022003002] = "kennedy center, NY"
}
You can also add a trailing separator after the last member.
The more detailed description of the table constructor can be found in 3.4.9 - Table Constructors. It could be summed up by the example provided there:
a = { [f(1)] = g; "x", "y"; x = 1, f(x), [30] = 23; 45 }
I really, really recommend using the Lua Reference Manual, it is an amazing helper.
I also highly encourage you to read some basic tutorials e.g. Learn Lua in 15 minutes. They should give you an overview of the language you are trying to use.
I would like to perform double substitution.
When printing:
def y = "\${x}"
def x = "world"
def z = "Hello ${y}"
println z
It prints:
Hello ${x}
When I would like it to print Hello World, I tried performing a double evaluation ${${}}, casting it to org.codehaus.groovy.runtime.GStringImpl, and a desperate ${y.toStrin()}
Edit:
To be more clear I mean this, but in Groovy:
https://unix.stackexchange.com/questions/68042/double-and-triple-substitution-in-bash-and-zsh
https://unix.stackexchange.com/questions/68035/foo-and-zsh
(Why I am doing this?: Because we have some text files that we need evaluate with groovy variables; the variables are many and in different part of the code are different, therefore I would like to have a solution working across all cases, not to have to bind each time each variable, not adding many lines of code)
So with what you have you're escaping the $ so it is just interpreted as a string.
For what you are looking to do I would look into Groovys's templating engines:
http://docs.groovy-lang.org/docs/next/html/documentation/template-engines.html
After reading your comment I played around with a few ideas and came up with this contrived answer, which is also probably not quite what you are looking for:
import groovy.lang.GroovyShell
class test{
String x = "world"
String y = "\${x}"
void function(){
GroovyShell shell = new GroovyShell();
Closure c = shell.evaluate("""{->"Hello $y"}""")
c.delegate = this
c.resolveStrategry = Closure.DELEGATE_FIRST
String z = c.call()
println z
}
}
new test().function()
But it was the closest thing I could come up with, and may lead you to something...
If I understand right, you are reading y from somewhere else. So you want to evaluate y as a GString after y and then x have been loaded. groovy.util.Eval will do this for simple cases. In this case, you have just one binding variable: x.
def y = '${x}'
def x = 'world'
def script = "Hello ${y}"
def z = Eval.me('x', x, '"' + script + '".toString()') // create a new GString expression from the string value of "script" and evaluate it to interpolate the value of "x"
println z
I'm trying to draw some boxes in Rascal and trying to give each box its own callback function. On entering the box with the mouse the corresponding string should get displayed in the text element (so hovering box1 should display box1 etc.).
However, at the moment the text does pop up but just displays "box3" for each of the 3 boxes.
Any ideas?
strings = ["box1", "box2", "box3"];
boxes = [ box(
size(100, 100),
onMouseEnter(void() {
output = s;
})
) | s <- strings];
render(hcat([
vcat(boxes),
text(str () {return output;})
]));
Good question, classical problem. The essence of the problem is that Rascal uses "non-capturing closures": this means that functions that are returned from another function share the same context. In your case this is the variable s introduced by s <- strings. This nearly always happens when you create function values in a loop (as you do here). The solution is to wrap another function layer around the returned function.
Here is a simple example:
list[int()] makeClosures()
= [ int() {return i;} | i <- [0,1,2]];
void wrong(){
lst = makeClosures();
println(lst[0]());
println(lst[1]());
println(lst[2]());
}
which will print surprisingly the values 2,2and2`. The solution is, as said, to introduce another function level:
int() makeClosure(int i)
= int() { return i;};
list[int()] makeClosuresOK()
= [ makeClosure(i) | i <- [0,1,2]];
void right(){
lst = makeClosuresOK();
println(lst[0]());
println(lst[1]());
println(lst[2]());
}
now calling right() will print 1, 2, and 3 as expected.
I leave it as an exercise how this is done in your example, but I am prepared to give a solution when you ask for it. Good luck!