What does left arrow <- mean outside a do block?

被刻印的时光 ゝ 提交于 2020-11-27 05:01:45

问题


I came across with the following code recently and it bothers me a lot

lowerSafeForeignCall dflags block
| (entry, middle, CmmForeignCall { .. }) <- blockSplit block
= do
 -- do block stuffs
 -- Block doesn't end in a safe foreign call:
| otherwise = return block

This piece of code is from https://phabricator.haskell.org/rGHCb0534f78a73f972e279eed4447a5687bd6a8308e

in file compiler/cmm/CmmLayoutStack.hs

line 983

I really would like to konw what is this <- in the second line. I believe lowerSafeForeignCall is a function and the | and 'otherwise' indicate this function uses guards. So

(entry, middle, CmmForeignCall { .. }) <- blockSplit block

must be of type Bool. But the <- is outside any do block. I did some search online but still not a single clue about this usage.


回答1:


That's a pattern guard:

guard       →   pat <- infixexp      (pattern guard)

[...]

A guard has one of the following forms:

  • pattern guards are of the form p <- e, where p is a pattern (see Section 3.17) of type t and e is an expression type t. They succeed if the expression e matches the pattern p, and introduce the bindings of the pattern to the environment.

Where normal guards are limited to a boolean check, pattern guards can match against an arbitrary pattern and define local variables. (In your case entry, middle, and the contents of CmmForeignCall will be directly available in the function body.)

You can think of boolean guards as equivalent to pattern guards with a pattern of True:

| expr

works like

| True <- expr



回答2:


This is a pattern guard [Haskell-wiki]. Since Haskell'10 a guard is a list of qualifiers. A qualifier can be a condition (like in the old guards), and pattern guards.

Haskell will thus (lazily) evaluate the expression on the right side of the arrow <- and aim to match it with the pattern on the left of the arrow. If that succeeds, than the guard (well that part of the guard) is successful. If all the parts of the guard are successful, then the rule "fires".

In this specific case the only part of the pattern that might fail is the fact that the third item of the 3-tuple is not a CmmForeignCall data constructor. Furthermore by using this pattern guard, we can of course use entry, middle in the body of the expression.



来源:https://stackoverflow.com/questions/56393471/what-does-left-arrow-mean-outside-a-do-block

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