Why does adding parentheses in if condition results in compile error?

Question

The following Go code runs OK:

package main

import "fmt"

func main() {
    if j := 9; j > 0 {
        fmt.Println(j)
    }
}

But after adding parentheses in condition:

package main

import "fmt"

func main() {
    if (j := 9; j > 0) {
        fmt.Println(j)
    }
}

There is compile error:

.\Hello.go:7: syntax error: unexpected :=, expecting )
.\Hello.go:11: syntax error: unexpected }

Why does the compiler complain about it?

Answer 1

The answer is not simply "because Go doesn't need parentheses"; see that the following example is a valid Go syntax:

j := 9
if (j > 0) {
    fmt.Println(j)
}

Go Spec: If statements:

IfStmt = "if" [ SimpleStmt ";" ] Expression Block [ "else" ( IfStmt | Block ) ] .

The difference between my example and yours is that my example only contains the Expression block. Expressions can be parenthesized as you want to (it will not be well formatted, but that is another question).

In your example you specified both the Simple statement and the Expression block. If you put the whole into parentheses, the compiler will try to interpret the whole as the Expression Block to which this does not qualify:

Expression = UnaryExpr | Expression binary_op UnaryExpr .

j > 0 is a valid expression, j := 9; j > 0 is not a valid expression.

Even j := 9 in itself is not an expression, it is a Short variable declaration. Moreover the simple assignment (e.g. j = 9) is not an expression in Go but a statement (Spec: Assignments). Note that assignment is usually an expression in other languages like C, Java etc.). This is the reason why for example the following code is also invalid:

x := 3
y := (x = 4)