s = space; l = line break; t = tab
ssstsssttsl
tl
ssl
sssttsssstl
tl
ssl
This only prints F and not a. I am not very certain about stacks and it is probably something to do with that.
I put your code into this online IDE: https://whitespace.kauaveel.ee/
It includes an on-the-fly disassembler, which is nice.
As is, your code produces an "Unexpected EOF" error. This is because of the last linebreak (l). Removing that gives
push 70
printc
label_0:
printc
from the disassembler. I.e. what you thought was your second push is actually a label instruction.
This is because the printc instruction is actually just tlss. You have an extra l after that, which combined with the following two spaces forms lss (label).
Fixed code:
s s s t s s s t t s l
t l
s s s s s t t s s s s t l
t l
s s
Related
I am very new to prolog, so assume that I know very little terminology.
I am using swipl in SWI-prolog.
I want to check if a string starts with a left squiggly bracket('{') and ends with a right squiggly bracket('}'}
Some answers that I have read online have lead me to program the following into my knowledge base to check if the string starts with a left squiggly bracket.
start_left_squiggle([Letter|_]):-
Letter = '{'.
But when I run this function, I get false, when I expect it to return true.
?- start_left_squiggle('{hello').
false.
As well, answers that seem correct for checking the if the last character is a squiggly bracket have lead me to code the following.
last_char(str, X):-
name(S, N),
reverse(N, [F|_]),
name(X, [F]).
end_right_squiggle(Werd):-
last_char(Werd, Last),
Last = '}'.
And I again get false when running the function, when I expect it to return true.
?- end_right_squiggle('hello}').
false.
Use sub_atom(Atom, Before, Length, After, Subatom) like so:
?- sub_atom('{abc}',0,1,_,C).
C = '{'.
?- sub_atom('{abc}',_,1,0,C).
C = '}'.
Or just test:
?- sub_atom('{abc}',0,1,_,'{').
true.
?- sub_atom('{abc}',_,1,0,'}').
true.
First thing you need to do is to break the atom into list of characters like this:
start_with_left(H):-
atom_chars(H,X), %here x is a list
X = [L|_], %get the head of the list which is frist element an compare
L == '{'.
You can use a recursive definition to check righ side of the atom after converting the atom into list of characters and when length of the list is 1 then compare it with bracket , it means if last element is same you should get true otherwise False.
Right is like this, it's same but we need last element so we have to use recursion:
start_with_right(H):-
atom_chars(H,X), %here x is a list
length(X,Y),
check_at_end(X,Y).
check_at_end([H|_],1):-
H == '}'.
check_at_end([_|T],Y):-
NewY is Y -1,
check_at_end(T,NewY).
.
I often read that I shouldn't mix tabs and spaces in Haskell, or that I shouldn't use tabs at all. Why?
The problem is twofold. First of all, Haskell is indentation sensitive, e.g. the following code isn't valid:
example = (a, b)
where
a = "Hello"
b = "World"
Both bindings need to be indented with the same number of spaces/tabs (see off-side rule). While it's obvious in this case, it's rather hidden in the following one, where I denote a space by · and a tab by »:
example = (a, b)
··where
····a = "Hello"
» b = "World"
This will look like valid Haskell code if the editor will show tabs aligned to multiples by four. But it isn't. Haskell tabs are aligned by multiples of eight, so the code will be interpreted like this:
example = (a, b)
··where
····a = "Hello"
» b = "World"
Second, if you use only tabs, you can end up with a layout that doesn't look right. For example, the following code looks correct if a tab gets displayed with six or more spaces (eight in this case):
example = (a, b)
» where» a = "Hello"
» » b = "World"
But in another editor that uses 4 spaces it won't look right anymore:
example = (a, b)
» where» a = "Hello"
» » b = "World"
It's still correct, though. However, someone who's used to spaces might reindent b' binding with spaces and end up with a parser error.
If you enforce a code convention throughout your code that makes sure that you only use tabs at the beginning of a line and use a newline after where, let or do you can avoid some of the problems (see 11). However, current releases of GHC warn about tabs by default, because they have been a source of many parser errors in the past, so you probably want to get rid of them too.
See also
A reddit thread on the topic (majority pro spaces, but some pro tabs)
Good Haskell Style (pro spaces)
Yet Another Tabs v Space debate (pro mixing)
I am using vim to log a lot of my day-to-day work. I usually have a lot of results with matrix. When I paste those at the end of the day I have pages of (sometimes) 0ne or 2 columns matrix
My question is how can I ask vim to programatically format
this:
R) DT = data.frame(x=rnorm(6),y=rnorm(6))
R) DT
x y
1 -0.0007499 0.7661
2 1.5559552 -0.6664
3 0.2082094 -0.1598
4 -0.1684766 -0.0536
5 3.1014233 -0.7439
6 0.1985104 2.0415
R) DT2 = data.frame(x=rnorm(7),y=rnorm(7))
R) DT2
x y
1 -0.005116 -0.4388
2 2.317381 0.6597
3 0.359883 0.4770
4 -1.426220 0.4911
5 0.468820 -0.3260
6 0.626147 -0.6254
7 -1.086660 2.1973
to this: side by side
R) DT = data.frame(x=rnorm(6),y=rnorm(6)) R) DT2 = data.frame(x=rnorm(7),y=rnorm(7))
R) DT R) DT2
x y x y
1 -0.0007499 0.7661 1 -0.005116 -0.4388
2 1.5559552 -0.6664 2 2.317381 0.6597
3 0.2082094 -0.1598 3 0.359883 0.4770
4 -0.1684766 -0.0536 4 -1.426220 0.4911
5 3.1014233 -0.7439 5 0.468820 -0.3260
6 0.1985104 2.0415 6 0.626147 -0.6254
7 -1.086660 2.1973
EDIT:
Thanks everybody, Kent solution worked, strangely before set ve=all my C-Q did not allow to select a block like
############
###############
#########
#########
#########
I couldn't reach the last column of the second line (using gvim on win XP).
Now works very well
I'll also use Ingo Karkat plugin"
try following steps:
:set ve=all
move cursor to beginning of line R) DT2 = data.frame(x=rnorm(7),y=rnorm(7))
press C-V, then using motion magic to select the 2nd block
press d
move cursor to the first line, the position you want to paste the 2nd block. type p
if your text is always like that format, you could try this mapping:
:nnoremap <leader>mt :let &ve='all'<cr>gg/^R)<CR>n<c-v>G$dgg$3lp
then you could in normal mode type <leader>mt to do the transformation.
NOTE
I have no experience of vim on windows... so you may have to change the mapping for windows blockwise selection. <C-Q> ? not sure.
I assume the first line of the file is R)...
this will paste the 2nd block to the position: 3 spaces after the end of 1st line, change the number 3 in mapping if you want to adjust it
The key to this is blockwise visual mode. Go to the beginning of the second block (what should become the right column next to the first block), press Ctrl+V (on Windows, this is often remapped to Ctrl+Q), create the selection with $ and jjj. . ., then delete with d. Then go to the first line of the first block, $ to go to the end of line, (optionally append padding whitespace), then paste with p . The second block should be pasted to the right of your cursor. Finally, you can delete the empty lines left where the second block was.
My UnconditionalPaste plugin simplifies this task: You can then simply delete the second block with dd or Vjjj. . .d, and force a paste in blockwise mode with the plugin's gdp mapping.
If you find a way to determine the size of the blocks (the } key, which jumps to the next empty line, may come handy here), you can then put all of the steps into a custom mapping.
I created a function to search a custom number of empty lines (with or without spaces) and to replace them with a new (custom) number of empty lines.
fun! s:AddRemoveNumbEmptyLines()
if !exists("emptylinesbefore")
let emptylinesbefore = "How many empty lines do you search? \n (p.e. 2,3 4, ,3)"
endif
let b = inputdialog(emptylinesbefore)
if !exists("emptylinesafter")
let emptylinesafter = "How many empty lines must it be?"
endif
let c = inputdialog(emptylinesafter)
let m = repeat('\r', c)
exe 's/\(^\s*$\n\)\{'.b.'}/'.m.'/gc'
endfun
Let say b = 2, (2 and more) AND m = 3
If vim finds 4 empty lines it does a substitution to 3 empty lines.
(thats ok).
But when I refuse the substitution request (I use the "c" (confirm) flag) it finds at the same place 3 empty lines and asks again if it has to be replaced with 3 empty lines. When I refuse again, it finds at the same place, 2 empty lines and asks again if I want to do a substitution.
How can I avoid these multiple substitution requests (at the same place)?
Hope I made myself clear :)
resolved it! I just had to check for a non space \S in the line before and the line after.
My new exe = 's/^.*\S\+.*\n\zs\(^\s*$\n\)\{'.b.'}\ze\s*\S\+/'.m.'/gc
I tried but the code wont work.
for. T do. B end.
for_xyz. T do. B end.
What would be the equivalent of this from C#
for(int i = 0; i < 10; i++)
Console.WriteLine("Hello World!");
And what's a good keyword to Google for J problems?
A more J-ish way to loop is using Power ^:, this
f^:10 y
will apply f 10 times; first to y, then to f(y), ... :
f(f(f(f(f(f(f(f(f(f(y))))))))))
So if p is a print function, eg: p =: (4) 1!:2~ ]:
(p^:10) 'Hello World!'
Hello World!Hello World!...
In general J (in a way) promotes loop-less code. If you really needed 10 times the string 'Hello World!' for example, you probably would do something like:
10 12 $ 'Hello World!'
Hello World!
Hello World!
Hello World!
...
As noted at the beginning of the Control Structures section, these only apply within Explicit definition. The colon is the key for setting up such a script. The only time 'for.' (or any similar word) can occur is within a script determined by the right parameter to : , i.e. colon, meaning Explicit.
Use the link on the control-word for. on that page to find complete samples. Notice that these special symbols (such as for. and end.) normally occur in multi-line scripts that end with a single lone right-paren. That sort of structure is what you must use if you're to use control words.
Here is the first of the examples given on the Dictionary page documenting the for. structure (http://jsoftware.com/help/dictionary/cfor.htm):
f0=. 3 : 0
s=. 0
for. i. y do. s=. >:s end.
)
Once you have arranged control words inside this sort of structure, they take effect when the script is executed. In this example, when the verb f0 receives an integer as its only (right) parameter (referred to as y in the script) it results in the same integer. It iterates through the for loop to arrive at that number.