5

In some source code I found this :

if etherbase != (common.Address{}) {
    return etherbase, nil
}

etherbase is of type common.Address and it is defined as this:

// Lengths of hashes and addresses in bytes.
const (
    HashLength    = 32
    AddressLength = 20
)
// Address represents the 20 byte address of an Ethereum account.
type Address [AddressLength]byte

The question is: what do parethesis mean in this context? Why can't they be omitted? Like this:

if etherbase != common.Address{} {
    return etherbase, nil
}
Nulik
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1 Answers1

10

The Go Programming Language Specification

Composite literals

A parsing ambiguity arises when a composite literal using the TypeName form of the LiteralType appears as an operand between the keyword and the opening brace of the block of an "if", "for", or "switch" statement, and the composite literal is not enclosed in parentheses, square brackets, or curly braces. In this rare case, the opening brace of the literal is erroneously parsed as the one introducing the block of statements. To resolve the ambiguity, the composite literal must appear within parentheses.

if x == (T{a,b,c}[i]) { … }
if (x == T{a,b,c}[i]) { … }

An ambiguous composite literal common.Address{} before an if block { … }.

if etherbase != common.Address{} {
    return etherbase, nil
}

An unambiguous composite literal (common.Address{}) before an if block { … }.

if etherbase != (common.Address{}) {
    return etherbase, nil
}

For example,

package main

const AddressLength = 20

type Address [AddressLength]byte

func f(etherbase Address) (Address, error) {
    // Unambiguous
    if etherbase != (Address{}) {
        return etherbase, nil
    }
    return Address{}, nil
}

func g(etherbase Address) (Address, error) {
    // Ambiguous
    if etherbase != Address{} {
        return etherbase, nil
    }
    return Address{}, nil
}

func main() {}

Playground: https://play.golang.org/p/G5-40eONgmD

peterSO
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