this one helps. Why doesn't Haskell allow more complex infix expressions, eg. , It can be faked code :
import Data.Function ( (&) )  = flip ($)
infixl 3
<,
>
(<) :: a > (a > b) > b
(<) = (&)
(>) :: (a > b) > (a > b)
(>) = ($)
as :: Num c => [c] > [c] > [c]
as xs ys = xs <zipWith (+)> ys
bs :: (Int > Int > a > b > c) > (a > b > c)
bs f l r = l <f 0 0> r
 >> cs
 [9,19,29,39,8,18,28,38]
cs :: [Int]
cs = [1..2] <^flip ()^> [10,20..40]
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Handling parenthesis while converting infix expressions to postfix expressions
By : Roe
Date : March 29 2020, 07:55 AM
hop of those help? You need to push the left parenthesis onto the stack, and process the stack like so when you encounter a right parenthesis: code :
// opening (
if (in_fix.peek().type == 4) {
post_fix.push(in_fix.pop());
}
//closing )
if(in_fix.peek().type == 5){
while(!(post_fix.isEmpty()  post_fix.peek().type == 4)){
postfixstr.append(post_fix.pop());
}
if (post_fix.isEmpty())
; // ERROR  unmatched )
else
post_fix.pop(); // pop the (
in_fix.pop(); // pop the )
}

Why does a compiler have trouble parsing infix expressions compared to postfix expressions?
By : MAC
Date : March 29 2020, 07:55 AM
Hope that helps Consider a + b * c / d e. According to conventional precedence rules, this should be parsed as a + ((b * c) / d). Simply reading from right to left would yield ((a + b) * c)/d, which is not what you want. The compiler needs to know the precendence (and associativity) of every operator and take them into account. On the other hand, postfix expressions have explicit precedence. For example, the first expression above is equivalent to b c * d / a +.

Regular Expressions to validade Prefix, Infix and Postfix expressions
By : harihara padhy
Date : March 29 2020, 07:55 AM
wish of those help Unfortunately this isn't possible without recursive regex (which C# doesn't support). While (wellformed) infix notation can be matched with the simple regex [AZ](?:[*/+][AZ])*, pre and postfix notation don't follow such a simple pattern that they could be matched by nonrecursive regex. That said, a very simple way to determine if an expression uses pre/postfix notation (without verifying its correctness) is to check whether the first/last character is an operator: ^[*/+] would match a wellformed prefix expression and .*[*/+]$ a wellformed postfix expression.

Parsing Infix Mathematical Expressions in Swift Using Regular Expressions
By : Jonathan Valiente
Date : March 29 2020, 07:55 AM
wish help you to fix your issue Your pattern does not work because it only matches text at the start of the string (see ^ anchor), then the (?=[+/*]) positive lookahead requires the first char to be an operator from the specified set but the only operator that you consume is an optional . So, when * tries to match the enclosed pattern sequence the second time with 99+44+2+3/3.26, it sees +44 and ?\d fails to match it (as it does not know how to match + with ?). Here is how your regex matches the string: code :
let pattern = "(?<!\\d)?\\d+(?:\\.\\d+)?[+*/%()]"
Optional(["31", "+", "2", "", "3", "*", "43.8", "/", "1", "%", "(", "1", "*", "2", ")"])

Do infix functions in Haskell have type expressions ?
By : electro_fantics
Date : March 29 2020, 07:55 AM
I wish did fix the issue. Yes, the infix functions are functions, but to use them with :t you need to write it as a normal prefix function: code :
:t (++)
(++) :: [a] > [a] > [a]
(+) 1 2
1 + 2
reverse `map` ["hello","world"]
map reverse ["hello","world"]

