Scala: Can you use "foo match { bar }" in an expression without parentheses? - scala

Question

Why are the parentheses needed here? Are there some precedence rules I should know?
scala> 'x' match { case _ => 1 } + 1
<console>:1: error: ';' expected but identifier found.
'x' match { case _ => 1 } + 1
^
scala> ('x' match { case _ => 1 }) + 1
res2: Int = 2
Thanks!

Answer 1

As Agilesteel says, a match is not considered as a simple expression, nor is an if statement, so you need to surround the expression with parentheses. From The Scala Language
Specification, 6 Expressions, p73, the match is an Expr, as is an if. Only SimpleExpr are accepted either side of the + operator.
To convert an Expr into a SimpleExpr, you have to surround it with ().
Copied for completeness:
Expr ::= (Bindings | id | ‘_’) ‘=>’ Expr
| Expr1
Expr1 ::= ‘if’ ‘(’ Expr ‘)’ {nl} Expr [[semi] else Expr]
| ‘while’ ‘(’ Expr ‘)’ {nl} Expr
| ‘try’ ‘{’ Block ‘}’ [‘catch’ ‘{’ CaseClauses ‘}’] [‘finally’ Expr]
| ‘do’ Expr [semi] ‘while’ ‘(’ Expr ’)’
| ‘for’ (‘(’ Enumerators ‘)’ | ‘{’ Enumerators ‘}’) {nl} [‘yield’] Expr
| ‘throw’ Expr
| ‘return’ [Expr]
| [SimpleExpr ‘.’] id ‘=’ Expr
| SimpleExpr1 ArgumentExprs ‘=’ Expr
| PostfixExpr
| PostfixExpr Ascription
| PostfixExpr ‘match’ ‘{’ CaseClauses ‘}’
PostfixExpr ::= InfixExpr [id [nl]]
InfixExpr ::= PrefixExpr
| InfixExpr id [nl] InfixExpr
PrefixExpr ::= [‘-’ | ‘+’ | ‘~’ | ‘!’] SimpleExpr
SimpleExpr ::= ‘new’ (ClassTemplate | TemplateBody)
| BlockExpr
| SimpleExpr1 [‘_’]
SimpleExpr1 ::= Literal
| Path
| ‘_’
| ‘(’ [Exprs] ‘)’
| SimpleExpr ‘.’ id s
| SimpleExpr TypeArgs
| SimpleExpr1 ArgumentExprs
| XmlExpr
Exprs ::= Expr {‘,’ Expr}
BlockExpr ::= ‘{’ CaseClauses ‘}’
| ‘{’ Block ‘}’
Block ::= {BlockStat semi} [ResultExpr]
ResultExpr ::= Expr1
| (Bindings | ([‘implicit’] id | ‘_’) ‘:’ CompoundType) ‘=>’ Block
Ascription ::= ‘:’ InfixType
| ‘:’ Annotation {Annotation}
| ‘:’ ‘_’ ‘*’

Answer 2

After some inspection in the Scala specification, I think I can give it a shot.
If I am wrong please correct me.
first, an if or match are defined as Expr - expressions.
You are trying to create an infix expression (defined by the use of the operator between two expressions)
However the especification (section 3.2.8) states that :
All type infix operators have the same precedence; parentheses have to
be used for grouping
It also also states that:
In a sequence of consecutive type infix operations t0 op1 t1 op2 . .
.opn tn, all operators op1, . . . , opn must have the same
associativity. If they are all left-associative, the sequence is
interpreted as (. . . (t0 op1 t1) op2 . . .) opn tn.
So my take is that Scala does not know what to reduce first: the match or the method '+' invocation.
Take a look at this answer
Please correct me if I am wrong.

Answer 3

A match expression is not considered as simple expression. Here is a similar example:
scala> val foo = "bar" + if(3 < 5) 3 else 5 // does not compile
scala> val foo = "bar" + (if(3 < 5) 3 else 5) // does compile
Apparently you can't write complex expressions wherever you want. I don't know why and hope that someone with more knowledge of the topic will give you a better answer.