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Unformatted text preview: TYPE INFERENCE AND LACKWIT GILBERT BERNSTEIN 1. TYPE CHECKING Im going to use a running example to make this easier to follow f : (int, int, int) -> int let f x y z = if x > 0 then x = y else x = z Suppose were compiling this code. The first thing wed do is parse it into an abstract syntax tree. Here is the AST for the func- tion f : AST PIC Notice at the beginning of the code snippet provided, there is a type signature: f : (int, int, int) -> int The goal of type checking is to ensure that the programmer is not lying when they make statements like f is a function that takes three int s and returns an int . Now suppose that someone gave us (totally out of the blue) a complete labeling of our AST with a type for every node. We might write down these labels like this TYPED AST PIC where x : int is a label on the expression x meaning that the ex- pression x has type int . Now our task of type checking is very sim- ple. All we need to do is make sure that all of these labels are con- sistent with each other, according to some set of typing rules . These typing rules are usually written like this: e 1 : bool e 2 : 1 e 3 : 1 if e 1 then e 2 else e 3 : 1 In english, we would read this rule as follows. Given that we know the following: that e 1 is an expression of type bool , that e 2 is an ex- pression of type 1 , and that e 3 is an expression of type 1 , then we 1 2 GILBERT BERNSTEIN can conclude that the compound expression if e 1 then e 2 else e 3 has type 1 . More importantly, notice that if we just flip this rule up- side down, superimpose it on our AST and substitute in the specific value int for the placeholder 1 then we get a perfect match with our labeling. In reality, we are not handed a complete labeling of our AST, so we have to generate it ourselves. We would start with our goal. To prove that the if-then-else statement has type int : PARTIAL AST PIC Then, we would reason in reverse that the sub-expressions must have the following types: PROPAGATED LABELING AST PIC Obviously, doing this gives us a simple recursive algorithm to type-check with. However, look what happens when we reach the leaf node x : BOOLEAN EXPRESSION X BRANCH ZOOM We need some way to be sure that x is of type int here, but in general x might be any type. To solve this problem we need to make some assumptions (to use Milnors language). In modern parlance,(to use Milnors language)....
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- Winter '11