Why is this panic example below a type error in golang?

user_1569557 0
Viewed 2761

Why is this panicf-sprintf causing a type error in Golang 1.11? Go doesn't explain the reason, even if it says this is a common mistake.

https://golang.org/doc/go1.11#vet

go vet is now enforced during the build.

func panicf(s string, i ...interface{}) { panic(fmt.Sprintf(s, i)) }

The test is returning

missing ... in args forwarded to printf-like function

vet describes this as

func (*ptrStringer) BadWrap(x int, args ...interface{}) string {
    return fmt.Sprint(args) // ERROR "missing ... in args forwarded to print-like function"
}

func (*ptrStringer) BadWrapf(x int, format string, args ...interface{}) string {
    return fmt.Sprintf(format, args) // ERROR "missing ... in args forwarded to printf-like function"

Please help explain ... in golang in this context.

Here is a functional playground: https://play.golang.org/p/DijjanQNkxK

go types
1 Answers

panicf() accepts i as a variadic, the same as fmt.Sprintf(). Therefore you have to tell the compiler that you want each value of i to be sent to fmt.Sprintf() instead of sending the entire thing as a slice.

So change the code to:

func panicf(s string, i ...interface{}) { panic(fmt.Sprintf(s, i...)) }
user_6756370
Related