Golang's Int flag and time.After

Question

I'm getting an invalid operation: *timeout * time.Second (mismatched types int and time.Duration) error when trying to run something similar to this

timeout := flag.Int("timeout", 30, "The time limit for answering questions.")
flag.Parse()
timeoutCh := time.After(*timeout * time.Second)

Just to be sure, I checked the type of *timeout using reflect.TypeOf() and it is in fact an int. But if I do timeoutCh := time.After(30 * time.Second) or use any other int value the code works.

What am I missing here?

Jake Burkhead · Accepted Answer · 2018-03-23T02:38:14.010

2

timeoutCh := time.After(time.Duration(*timeout) * time.Second)

You have to convert *timeout which is type int to type time.Duration. The reason why time.After(30 * time.Second) works is that 30 is untyped and is converted to the type of time.Second which is time.Duration. See https://golang.org/ref/spec#Operators. Similarly this code works

x := uint(42)
if x == 42 {
    println("works!")
}

but this code won't compile

x := uint(42)
y := 42 // defaults to type int
if x == y {
    println("this won't complile!")
}

edited Mar 23 '18 at 02:38

answered Mar 23 '18 at 02:30

Jake Burkhead

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"Except for shift operations, if one operand is an untyped constant and the other operand is not, the constant is converted to the type of the other operand." I didn't know that! Thank you! – Wilder Galvao Mar 24 '18 at 03:19

score 2 · Answer 2 · edited Mar 23 '18 at 03:12

You cannot multiply two different types, so you will need to convert the integer to a time.Duration type. You can do that by simply casting it like this:

time.Duration(*timeout)

A "unit" of time is technically a nanosecond and time.Second is one seconds worth of nanoseconds. The math works out though so you can simply say something like this for 3 seconds:

time.Duration(3) * time.Second

Golang's Int flag and time.After

2 Answers2