let x = 2;;
let x = 3
in let y = x +1
in x + 1;;
let x = 3 and y = x + 1
in x + y;;
The results are ordered respectively to each expression:
val x : int = 2
- : int = 7
- : int = 6
Can you help me understand why the last expression equals 6, it's hard for me to understand it.
I run this code in tryOCamlpro.
The first one is self-explanatory.
The second one should warn you that y is unused. x is 3, so x + 1 is 4. Your observed result of 7 suggests that the code you've shown us is not actually what you evaluated.
For the third, let's try something a bit different:
# let x = 3 and y = x + 1
in (x, y);;
- : int * int = (3, 3)
Huh. Let's fire up a clean interpreter environment and try that:
# let x = 3 and y = x + 1 in (x, y);;
Error: Unbound value x
From this we can see that the definition of y wasn't using the x defined as 3 but rather the previous definition of x as 2.
You likely wanted:
# let x = 3 in
let y = x + 1 in
x + y;;
- : int = 7
When you use let <pattern> = <expr> and <pattern> = <expr> or let <pattern> = <expr> and <pattern> = <expr> in <expr> the definitions are independent of one another. As a result, this is much less common than let <pattern> = <expr> let <pattern> = <expr> or let <pattern> = <expr> let <pattern> = <expr> in <expr>
If we modify this construct to let rec <pattern> = <expr> and <pattern> = <expr> or let rec <pattern> = <expr> and <pattern> = <expr> in <expr> we can define mutually recursive functions. This is not common.