问题
I want to give a tuple to a printf function:
let tuple = ("Hello", "world")
do printfn "%s %s" tuple
This, of course, does not work, compiler first says, that it needs string instead of string*string. I write it as follows:
let tuple = ("Hello", "world")
do printfn "%s %s" <| fst tuple
Then compiler reasonably notes that now I have function value of type string -> unit. Makes sense. I can write
let tuple = ("Hello", "world")
do printfn "%s %s" <| fst tuple <| snd tuple
And it works for me. But I'm wondering, if there might be any way to do it nicer, like
let tuple = ("Hello", "world")
do printfn "%s %s" <| magic tuple
My problem is that I can't get which type does printf need so that to print two arguments.
What could magic function look like?
回答1:
You want
let tuple = ("Hello", "world")
printfn "%s %s" <|| tuple
Notice the double || in <|| and not a single | in <|
See: MSDN <||
You can also do
let tuple = ("Hello", "world")
tuple
||> printfn "%s %s"
There are other similar operators such as |>, ||>, |||>, <|, <||, and <|||.
A idiomatic way to do it using fst and snd is
let tuple = ("Hello", "world")
printfn "%s %s" (fst tuple) (snd tuple)
The reason you don't usually see a tuple passed to a function with one of the ||> or <|| operators is because of what is known as a destructuring.
A destructing expression takes a compound type and destructs it into parts.
So for the tuple ("Hello", "world") we can create a destructor which breaks the tuple into two parts.
let (a,b) = tuple
I know this may look like a tuple constructor to someone new to F#, or may look even odder because we have two values being bound to, (noticed I said bound and not assigned), but it takes the tuple with two values and destructured it into two separate values.
So here we do it using a destructuring expression.
let tuple = ("Hello", "world")
let (a,b) = tuple
printfn "%s %s" a b
or more commonly
let (a,b) = ("Hello", "world")
printfn "%s %s" a b
来源：https://stackoverflow.com/questions/20568309/feeding-tuple-into-function-such-as-printfn