问题
I am writing my language in Pegjs and as usual, my language has some keywords, like true
, false
, if
, else
and today
for instance. Now, I want to declare a variable, but apparently, the variable name cannot be one of the reserved keywords. It can be any alpha followed by an alpha-numeric, with the exception of the language keywords.
I did the following (testable in Pegjs Online):
variable = c:(alpha alphanum*)
{
var keywords = ["true", "false", "if", "else", "today"];
var res = c[0]
for (var i = 0; i<c[1].length; i++) {
res=res+c[1][i]
}
if(keywords.indexOf(res)>=0) {
return error('\'' + res + '\'' + ' is a keyword and cannot be used as a variable name.');
}
return { 'dataType' : 'variable', 'dataValue' : res };
}
alpha = [a-zA-Z]
alphanum = [a-zA-Z0-9_]
boolean = v: ("true" / "false")
{
return { 'dataType' : 'boolean', 'dataValue': v};
}
Now true
is illegal, but true1
is not. This is fine. However, since I have defined the boolean
structure somewhere else in my language, is it not possible to re-use that definition instead of manually re-defining the non-allowed keywords inside my variable
definition?
You can imagine why my solution is error-prone. I tried several things but they did not work.
Thanks for your help!
回答1:
Simple Answer:
(See this code in action at http://peg.arcanis.fr/2VbQ5G/)
variable = ! keyword (alpha alphanum*)
{
return { 'dataType' : 'variable', 'dataValue': text()};
}
keyword = "true" / "false" / "if" / "else" / "today"
alpha = [a-zA-Z]
alphanum = [a-zA-Z0-9_]
boolean = ("true" / "false")
{
return { 'dataType' : 'boolean', 'dataValue': text()};
}
Note: This loses your helpful error reporting. If I get a chance, I'll try to put up an answer that retains it.
The important bit of the code below is at the beginning of the variable
rule: ! keyword
. This easiest way to grok this as that the parser is looking ahead 1 token. If what it finds is not a keyword, then it allows the rule to try to match the token. If, on the other hand, it is a keyword, then the ! keyword
expression (and by extension, the whole variable
rule fails.
To quote David Majda's documentation:
! expression
Try to match the expression. If the match does not succeed, just return undefined and do not advance the parser position, otherwise consider the match failed.
来源:https://stackoverflow.com/questions/26976820/pegjs-dont-allow-reserved-keywords-as-a-variable-name