Printing a Go type ignoring "String() string" method

Question

If I have a type in go like this:

type myType ...

func (m myType) String() string { ... }

how can I print (using the various fmt functions) this type using the default representation (that is, instead of having String() called)? What I'd like to do is something like this:

func (m myType) String() string {
    // some arbitrary property
    if myType.isValid() {
        // format properly
    } else {
        // will recurse infinitely; would like default
        // representation instead
        return fmt.Sprintf("invalid myType: %v", m)
    }
}

Mr_Pink · Accepted Answer · 2014-09-24 15:11:28Z

6

fmt.Stringer is the default format, which is called when you use %v. If you want the Go syntax, use %#v.

Alternatively, you can bypass the reflection in fmt altogether, and format your output as you see fit.

func (m myType) String() string {
    return fmt.Sprintf("{Field: %s}", m.Value)
}

If the underlying type of myType is a number, string or other simple type, then convert to the underlying type when printing:

func (m mType) String() string {
    return fmt.Sprint(int(m))
}

edited Sep 24, 2014 at 15:11

answered Sep 24, 2014 at 14:39

Mr_Pink

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Comments

nos · Accepted Answer · 2014-09-24 14:37:31Z

5

Use %#v instead of %v

That will not invoke String(). - but it will invoke GoString() if you implement it.

answered Sep 24, 2014 at 14:37

nos

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2 Comments

Ainar-G Over a year ago

This was covered in Effective Go. OP should convert the value to the underlying type and then use %v format. See my answer.

nos Over a year ago

@Ainar-G Your answer is trivial for a type alias, but doesn't become straight forward for e.g. type myType struct { a myOtherType}. There also doesn't seem to be anything in Effective Go that indicates %#v is not what the OP wants.

Ainar-G · Accepted Answer · 2014-09-24 15:17:32Z

2

Using %#v format is not the correct answer if you want to get your underlying type's String to work or your type is a type alias.

As is explained in Effective Go, just convert it back to the type it represents:

type Foo int

func (f Foo) String() string {
    if f == 0 {
        return "foo"
    }
    return fmt.Sprintf("%v", int(f)) // N.B.
}

func main() {
    fmt.Println(Foo(0))
    fmt.Println(Foo(42))
}

Playground.

EDIT: As others have pointed out in the comments, if your type is a struct, using %#v format seems like the only way besides converting it to an anonymous struct type with same fields.

edited Sep 24, 2014 at 15:17

answered Sep 24, 2014 at 14:42

Ainar-G

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2 Comments

Mr_Pink Over a year ago

This only really helps though if myType is a type alias. More frequently it's a struct itself.

joshlf Over a year ago

Is there any way to do this that doesn't involve explicit conversion? My use case is that I have a struct, and it'd be a pain to create an anonymous version of it just for printing and then convert between the two.

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