I just set up coffeelint on Atom. I was getting plenty of warnings like this one:
if HelpersValidators.isNotEmpty(email) and HelpersValidators.isNotEmpty(password) and HelpersValidators.isEmail(email) and HelpersValidators.areValidPasswords(password, passwordConfirm)
...
=> Line exceeds maximum allowed length
So what I did is:
if HelpersValidators.isNotEmpty(email)
and HelpersValidators.isNotEmpty(password)
and HelpersValidators.isEmail(email)
and HelpersValidators.areValidPasswords(password, passwordConfirm)
But now, I'm getting the following error and I can't resolve it.
=> SyntaxError: unexpected LOGIC
So how can I structure this condition to make it appear on several lines?
You can't start new line with logic operator in CoffeeScript, so you should fold the line afret and operator, not before it:
if HelpersValidators.isNotEmpty(email) and
HelpersValidators.isNotEmpty(password) and
HelpersValidators.isEmail(email) and
HelpersValidators.areValidPasswords(password, passwordConfirm)
// do something
or
if HelpersValidators.isNotEmpty(email) and
HelpersValidators.isNotEmpty(password) and
HelpersValidators.isEmail(email) and
HelpersValidators.areValidPasswords(password, passwordConfirm)
// do something
Related
function check(str,arg;type=DataType,max=nothing,min=nothing,description="")
#argcheck typeof(arg)==type
#argcheck arg>min
#argcheck arg<max
#argcheck typeof(description)==String
return arg
end
function constr(name,arg,field)
return :(function $name($arg,$field)
new(check($name,$arg,$field))
end)
end
macro creatStruct(name,arg)
code = Base.remove_linenums!(quote
struct $name
end
end)
print(arg)
append!(code.args[1].args[3].args,[constr(name,arg.args[1].args[1],arg.args[1].args[2])])
code
end
macro myStruct(name,arg)
#creatStruct name arg
end
#myStruct test12 (
(arg1,(max=10))
)
In my code above I'm trying to build a macro that Creates a struct, and within the struct, you can define an argument with boundaries (max, min) and description, etc.
I'm getting this error:
syntax: "#141#max = 10" is not a valid function argument name
and every time I'm trying to solve it, I get another error like:
LoadError: syntax: "struct" expression not at top level
So, I think my Code/Approach is not that cohesive. Anybody can suggest tips and/or another Approche.
You're attempting to make an argument name max with a default value of 10. The error is about max=10 not being a valid name (Symbol), while max is. The bigger issue is you're trying to put this in the struct expression instead of a constructor method:
struct Foo
bar::Float64
max::Int64
end
# constructor
Foo(bar, max=10) = Foo(bar, max)
So you have to figure out how to make an expression for a method with default values, too.
Your second error means that structs must be defined in the top-level. "Top-level" is like global scope but stricter in some contexts; I don't know the exact difference, but it definitely excludes local scopes (macro, function, etc). It looks like the issue is the expression returned by creatStruct being evaluated as code in myStruct, but the LoadError I'm getting has a different message. In any case, the error goes away if I make sure things stay as expressions:
macro myStruct(name,arg)
:(#creatStruct $name $arg)
end
my goal is change a function to a format where the return value of the function is treated :
For example ; treating a the function scanf()
Return value of scanf : The value EOF is returned if the end of input is reached before
either the first successful conversion or a matching failure occurs.
EOF is also returned if a read error occurs, in which case the error
indicator for the stream (see ferror(3)) is set, and errno is set to
indicate the error.
Thus
scanf("%d\n",&i);
will be change to
#define RVAL(exp) do {if ((exp) == -1) { perror (#exp); exit(1); }} while (0)
...
RVAL(scanf("%d\n",&i));
Thus I want this to be done quickly means :
so what i do is look for occurences of "scanf" and replace it with "RVAL(scanf"
but the problem is i have to add another right parentheses
Can this be done fast ? using a techniques ? or a style ? where each whenever I enter scanf(); its replaced witch rval(scanf());
If you don't have many ) in your format string you can use a regex with (scanf([^)]*)); and replace with rval(\1);
*see comment
I try to check if input argument is of specific type and throw error message like:
function test(input)
if ~ischar(input)
error('%s is invalid input type.', class(input));
end
end
But Matlab shows error message with backtrace-information:
>> test(1)
Error using test (line 3)
double is invalid input type.
How can I turn off the line Error using test (line 3)?
I'm looking for something similar to off backtrace with warning: warning off backtrace;.
I'm not sure you can. The closest I got was by defining my own error structure:
testerr.message = 'test';
testerr.identifier = '';
testerr.stack.file = '';
testerr.stack.name = 'Test Thing';
testerr.stack.line = 1;
error(testerr)
Which returns:
Error using Test Thing
test
As long as you keep the file field blank it will not display the line specified in the stack.
One potential workaround could be a combination of fprintf and return, courtesy of Undocumented MATLAB:
function test(input)
if ~ischar(input)
fprintf(2, '%s is invalid input type.\n', class(input));
return
end
end
Depending on where this check resides in your real function you might need to get creative with how it exits, since return only kicks you back to the invoking function. Probably have it output a True/False flag?
I am very new to EPL queries.
Wrote this and it is throwing syntax error.
#Name('ExpressionTotalQuantitySoFar')
#Description('Gets the total quantity of a symbol so far')
create expression totalQuantitySoFar{ (TAX) =>
(Select sum(T.quantity) from TaxlotWindow as T where T.symbol = TAX.symbol and T.taxlotId < TAX.taxlotId)
};
create variable double totQty = 5.0 ;
#Name('ExpressionLongDebitBalanceTaxlotNoBox')
#Description('Check is if a trade side is invalid, returns rue for invalid statements')
create expression longDebitBalanceTaxlotNoBox{ (SECUR,TAX,ORD,AUE,FX) =>
totQty = totalQuantitySoFar(TAX)
case when (totQty > 0)
then cashImpactBase(SECUR,TAX,ORD,AUE,FX)*(-1)
else
0.0
end
};
It gives syntax error near case.
Any help?
Always include the syntax error text when posting. Else how is one supposed to be able to help.
My tip would be to simplify until the syntax is fine. Then add back stuff.
Most likely this strange declaration "totQty=.." is the cause as its wrong. EPL expressions are not a programming language and don't allow variable declarations like in Java or Scala. Perhaps just use a Java static method to compute instead of you need a programming language.
On compiling the following code with Scala 2.7.3,
package spoj
object Prime1 {
def main(args: Array[String]) {
def isPrime(n: Int) = (n != 1) && (2 to n/2 forall (n % _ != 0))
val read = new java.util.Scanner(System.in)
var nTests = read nextInt // [*]
while(nTests > 0) {
val (start, end) = (read nextInt, read nextInt)
start to end filter(isPrime(_)) foreach println
println
nTests -= 1
}
}
}
I get the following compile time error :
PRIME1.scala:8: error: illegal start of simple expression
while(nTests > 0) {
^
PRIME1.scala:14: error: block must end in result expression, not in definition
}
^
two errors found
When I add a semicolon at the end of the line commented as [*], the program compiles fine. Can anyone please explain why does Scala's semicolon inference fail to work on that particular line?
Is it because scala is assuming that you are using the syntax a foo b (equivalent to a.foo(b)) in your call to readInt. That is, it assumes that the while loop is the argument to readInt (recall that every expression has a type) and hence the last statement is a declaration:
var ntests = read nextInt x
wherex is your while block.
I must say that, as a point of preference, I've now returned to using the usual a.foo(b) syntax over a foo b unless specifically working with a DSL which was designed with that use in mind (like actors' a ! b). It makes things much clearer in general and you don't get bitten by weird stuff like this!
Additional comment to the answer by oxbow_lakes...
var ntests = read nextInt()
Should fix things for you as an alternative to the semicolon
To add a little more about the semicolon inference, Scala actually does this in two stages. First it infers a special token called nl by the language spec. The parser allows nl to be used as a statement separator, as well as semicolons. However, nl is also permitted in a few other places by the grammar. In particular, a single nl is allowed after infix operators when the first token on the next line can start an expression -- and while can start an expression, which is why it interprets it that way. Unfortunately, although while can start a expression, a while statement cannot be used in an infix expression, hence the error. Personally, it seems a rather quirky way for the parser to work, but there's quite plausibly a sane rationale behind it for all I know!
As yet another option to the others suggested, putting a blank newline between your [*] line and the while line will also fix the problem, because only a single nl is permitted after infix operators, so multiple nls forces a different interpretation by the parser.