I have 2 shift reduces when writting this sentence in yacc:
%%
%token blablabla
%%
expresion: expresion + expresion { ; }
|/*lambda*/
;
Does anyone know how to delete the warnings shift-reduce? I looked over this link but i got bison errors and did not work http://nereida.deioc.ull.es/~pl/perlexamples/node160.html
You can use
%left '+'
to tell YACC how to resolve the shift reduce conflict. That's the best way to do it.
You can make the grammar unambiguous by making a different term
expression = lower_expression + expression ;
That method gets tedious when there are a lot of operators and different levels of precedence.
Related
To get the path of a currently running script people suggest to use ${(%):-%N}. I have looked into this and I cannot wrap my head around why ${(%)%N} shouldn't work.
From reading the manual (zshexpn) it seems like the above command uses ${name:-value} which returns the value of name if this is set (and non-null), otherwise value. In the case above name is left blank, in which case value is always returned.
The first (%) is a parameter expansion flag that (as far as I understand) should make all % escapes (%N in this case) expand in the same way as in the prompt.
Shouldn't it be possible to simply use ${(%)%N} instead? This does not work, but I do not understand why.
Can anyone shed any light on this? Perhaps my understanding of how ${(%):-%N} is parsed is not correct?
The syntax ${...} is for a parameter expansion, and in this case %N is not a parameter - it's a string literal. The :- syntax is a way to turn a string literal into a parameter.
Some examples that may help explain this:
> parm='%N'
> print ${parm}
%N
> print -- ${(%)parm}
-zsh
> print X${notaparameter}X
XX
> print X${%N}X
XX
> print X${:-%N}X
X%NX
> print X${(%):-%N}X
X-zshX
I work with knitr() and I wish to transform inline Latex commands like "\label" and "\ref", depending on the output target (Latex or HTML).
In order to do that, I need to (programmatically) generate valid R strings that correctly represent the backslash: for example "\label" should become "\\label". The goal would be to replace all backslashes in a text fragment with double-backslashes.
but it seems that I cannot even read these strings, let alone process them: if I define:
okstr <- function(str) "do something"
then when I call
okstr("\label")
I directly get an error "unrecognized escape sequence"
(of course, as \l is faultly)
So my question is : does anybody know a way to read strings (in R), without using the escaping mechanism ?
Yes, I know I could do it manually, but that's the point: I need to do it programmatically.
There are many questions that are close to this one, and I have spent some time browsing, but I have found none that yields a workable solution for this.
Best regards.
Inside R code, you need to adhere to R’s syntactic conventions. And since \ in strings is used as an escape character, it needs to form a valid escape sequence (and \l isn’t a valid escape sequence in R).
There is simply no way around this.
But if you are reading the string from elsewhere, e.g. using readLines, scan or any of the other file reading functions, you are already getting the correct string, and no handling is necessary.
Alternatively, if you absolutely want to write LaTeX-like commands in literal strings inside R, just use a different character for \; for instance, +. Just make sure that your function correctly handles it everywhere, and that you keep a way of getting a literal + back. Here’s a suggestion:
okstr("+label{1 ++ 2}")
The implementation of okstr then needs to replace single + by \, and double ++ by + (making the above result in \label{1 + 2}). But consider in which order this needs to happen, and how you’d like to treat more complex cases; for instance, what should the following yield: okstr("1 +++label")?
I have the following in my .zshrc:
setopt PROMPT_SUBST
precmd(){
echo""
LEFT="$fg[cyan]$USERNAME#$HOST $fg[green]$PWD"
RIGHT="$fg[yellow]$(date +%I:%M\ %P)"
RIGHTWIDTH=$(($COLUMNS-${#LEFT}))
echo $LEFT${(l:$RIGHTWIDTH:)RIGHT}
}
PROMPT="$ "
This gives me the following screenshot
The time part on the right is always not going all the way to the edge of the terminal, even when resized. I think this is due to the $(date +%I:%M\ %P) Anyone know how to fix this?
EDIT: Zoomed in screenshot
While your idea is commendable, the problem you suffer from is that your LEFT and RIGHT contains ANSI escape sequences (for colors), which should be zero-width characters, but are nevertheless counted toward the length of a string if you naively use $#name, or ${(l:expr:)name}.
Also, as a matter of style, you're better off using Zsh's builtin prompt expansion, which wraps a lot of common things people may want to see in their prompts in short percent escape sequences. In particular, there are builtin sequences for colors, so you don't need to rely on nonstandard $fg[blah].
Below is an approximate of your prompt written in my preferred coding style... Not exactly, I made everything super verbose so as to be understandable (hopefully). The lengths of left and right preprompts are calculated after stripping the escape sequences for colors and performing prompt expansion, which gives the correct display length (I can't possibly whip that up in minutes; I ripped the expression off pure).
precmd(){
local preprompt_left="%F{cyan}%n#%m %F{green}%~"
local preprompt_right="%F{yellow}%D{%I:%M %p}%f"
local preprompt_left_length=${#${(S%%)preprompt_left//(\%([KF1]|)\{*\}|\%[Bbkf])}}
local preprompt_right_length=${#${(S%%)preprompt_right//(\%([KF1]|)\{*\}|\%[Bbkf])}}
local num_filler_spaces=$((COLUMNS - preprompt_left_length - preprompt_right_length))
print -Pr $'\n'"$preprompt_left${(l:$num_filler_spaces:)}$preprompt_right"
}
PROMPT="$ "
Edit: In some terminal emulators, printing exactly $COLUMN characters might wrap the line. In that case, replace the appropriate line with
local num_filler_spaces=$((COLUMNS - preprompt_left_length - preprompt_right_length - 1))
End of edit.
This is very customizable, because you can put almost anything in preprompt_left and preprompt_right and still get the correct lengths — just remember to use prompt escape sequence for zero width sequences, e.g., %F{}%f for colors, %B%b for bold, etc. Again, read the docs on prompt expansion: http://zsh.sourceforge.net/Doc/Release/Prompt-Expansion.html.
Note: You might notice that %D{%I:%M %p} expands to things like 11:35 PM. That's because I would like to use %P to get pm, but not every implementation of strftime supports %P. Worst case scenario: if you really want lowercase but %P is not supported, use your original command subsitution $(date +'%I:%M %P').
Also, I'm using %~ instead of %/, so you'll get ~/Desktop instead of /c/Users/johndoe/Desktop. Some like it, some don't. However, as I said, this is easily customizable.
I was going through hoc program written in The Unix Programming Environment. Came across the grammar as
list: /* nothing */
| list '\n'
| list expr '\n' { printf("\t%.8g\n", $2); }
;
What is the need of
| list '\n'
Looks like it will work even if we dont give that part of grammar! Then what is the importance of that part?
Printf is given with %.8g. What does that mean?
What to do if I need to give a prompt symbol? Just like $ in shell ?
Three completely unrelated questions here
The rule list: list '\n' allows it to accept (and ignore) blank lines in the input. Without it, any blank line would cause a syntax error.
The format specifier %.8g mean print a double argument with 8 digits of precision, using either normal or scientific notation, depending on what the exponent is.
If you want to print a prompt symbol, the easiest way is to do it in the lexer -- each time your yylex function reads a newline, it should print the prompt for the next line before returning the newline to the parser. If you're using flex, this could be a rule like:
\n { if (interactive) {
printf("$ ");
fflush(stdout); }
return '\n'; }
you might want to check if your input is coming from a terminal or file and set the global interactive flag appropriately.
I'm trying to learn how to configure my .vimrc file with my own functions.
I'd like to write a function that traverses every line in a file and counts the total number of characters, but ignores all whitespace. This is for a programming exercise and as a stepping stone to more complex programs (I know there are other ways to get this example value using Vim or external programs).
Here's what I have so far:
function countchars()
let line = 0
let count = 0
while line < line("$")
" update count here, don't count whitespace
let line = getline(".")
return count
endfun
What functional code could I replace that commented line with?
If I understand the question correctly, you're looking to count the number of non-whitespace characters in a line. A fairly simple way to do this is to remove the whitespace and look at the length of the resulting line. Therefore, something like this:
function! Countchars()
let l = 1
let char_count = 0
while l <= line("$")
let char_count += len(substitute(getline(l), '\s', '', 'g'))
let l += 1
endwhile
return char_count
endfunction
The key part of the answer to the question is the use of substitute. The command is:
substitute(expr,pattern,repl,flags)
expr in this case is getline(l) where l is the number of the line being iterated over. getline() returns the content of the line, so this is what is being parsed. The pattern is the regular expression \s which matches any single whitespace character. It is replaced with '', i.e. an empty string. The flag g makes it repeat the substitute as many times as whitespace is found on the line.
Once the substitution is complete, len() gives the number of non-whitespace characters and this is added to the current value of char_count with +=.
A few things that I've changed from your sample:
The function name starts with a capital letter (this is a requirement for user defined functions: see :help user-functions)
I've renamed count to char_count as you can't have a variable with the same name as a function and count() is a built-in function
Likewise for line: I renamed this to l
The first line in a file is line 1, not line 0, so I initialised l to 1
The while loop counted up to but not including the last line, I assume you wanted all the lines in the file (this is probably related to the line numbering starting at 1): I changed your code to use <= instead of <
Blocks aren't based on indentation in vim, so the while needs an endwhile
In your function, you have let line = getline('.')
I added a ! on the function definition as it makes incremental development much easier (everytime you re-source the file, it will override the function with the new version rather than spitting out an error message about it already existing).
Incrementing through the file works slightly differently...
In your function, you had let line = getline('.'). Ignoring the variable name, there are still some problems with this implementation. I think what you meant was let l = line('.'), which gives the line number of the current line. getline('.') gives the contents of the current line, so the comparison on the while line would be comparing the content of the current line with the number of the last line and this would fail. The other problem is that you're not actually moving through the file, so the current line would be whichever line you were on when you called the function and would never change, resulting in an infinite loop. I've replaced this with a simple += 1 to step through the file.
There are ways in which the current line would be a useful way to do this, for example if you were writing a function with that took a range of lines, but I think I've written enough for now and the above will hopefully get you going for now. There are plenty of people on stackoverflow to help with any issues anyway!
Have a look at:
:help usr_41.txt
:help function-list
:help user-functions
:help substitute()
along with the :help followed by the various things I used in the function (getline(), line(), let+= etc).
Hope that was helpful.
This approach uses lists:
function! Countchars()
let n = 0
for line in getline(1,line('$'))
let n += len(split(line,'\zs\s*'))
endfor
return n
endfunction
I suppose you have already found the solution.
Just for info:
I use this to count characters without spaces in Vim:
%s/\S/&/gn