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An if, semicolon, and let gotcha

tl;dr

When a let expression begins in the final branch of an if expression, a semicolon will not terminate the let; instead the let expression will include the code that comes after the semicolon. One might expect that the semicolon ends the if expression, and that the code following it will execute after the entire if expression, but instead that code is part of the if expression. This can introduce subtle bugs because the code may behave as expected when the final branch of the if executes, but not when it doesn’t. Indentation that suggests that the code after the semicolon is not part of the if and let can also make the bug difficult to see. To make the semicolon follow the entire if with the embedded let, wrap the let expression in parentheses or begin/end. (The apparent problem is a consequence of the normal, useful behavior of let interacting user expectations for how if works.)

The problem

Suppose we have an if expression executed for the sake of side effects. This is a complete expression. A subsequent semicolon will sequence the next expression so that it is always executed:

let foo n = 
  if n < 0 
  then print_string "low\n" 
  else print_string "high\n";
  print_string "ok\n"

Let’s try it out in utop:

# foo 42;;
high
ok
- : unit = ()
# foo (-42);;
low
ok
- : unit = ()

So far, so good. Now we decide to introduce a let inside the first, then branch of the if, like this:

let bar x =
  if x < 0
  then let message = "low\n" in
    print_string message
  else print_string "high\n";
  print_string "ok\n"

This behaves identically to the foo function defined above. Great.

Next we define a function in which the let is in the second, else, branch of the if:

let buggy_baz x =
  if x < 0
  then print_string "low\n"
  else let message = "high\n" in
    print_string message;
  print_string "ok\n"

Let’s try it out in utop:

# buggy_baz 42;;
high
ok
- : unit = ()
# buggy_baz (-42);;
low
- : unit = ()

Why is the final “ok” output missing in the second example? This isn’t what we intended.

This problem will also occur with a single-branch if/then expression and a let in the then branch.

Why does the problem occur?

The problem is that the let expression captured the final print_string. OCaml interpreted the semicolon before print_string "ok\n" as sequencing it within the scope of the let expression, rather than after the entire if/then/else. As a result, print_string "ok\n" only runs when the else clause is executed.

Misleading indentation, as in the example above, can make it difficult to see the problem. print_string "ok\n" should be indented in the same way as the line above it.

This behavior is really just a consequence of the normal (and useful) scoping rule for let. Here’s an example in utop:

# let x = 3 in
print_int x;
print_string ",";
print_int x;
print_string "\n";;
3,3
- : unit = ()

let is designed have a scope that extends past semicolons. (How far? That’s a bigger question about OCaml syntax more generally.) The problem is that when the let is in the last branch of the if, it can be natural for us to think that the semicolon ends the if expression, as it would if the let wasn’t part of the branch. However, for if, let and semicolon to behave that way would require let to have a different behavior when it was placed inside an if expression.

The solution

Misleading indentation can be avoided by using an in-editor code formatting tool such as ocp-indent. See Editor Tools on the Code Tools page.

We can make the code after the semicolon execute after the if and let expressions by explicitly delimiting the scope of the inner let using parentheses or begin/end:

let baz x =
  if x < 0
  then print_string "low\n"
  else (let message = "high\n" in
        print_string message);
  print_string "ok\n"

Or:

let baz x =
  if x < 0
  then print_string "low\n"
  else begin 
    let message = "high\n" in
    print_string message
  end;
  print_string "ok\n"

These functions will behave identically to the original foo function above. Another solution is to wrap the entire if expression in parentheses or begin/end.

Note that you have to include a semicolon after the closing ) or end. Otherwise you will get a potentially confusing error message such as this one:

Error: This expression has type unit
       This is not a function; it cannot be applied.

Summary

The general rule is that let will capture anything that’s sequenced in whatever happens to be the last if clause. This is because the scope of let always extends as far as it can into subsequent semicolon-delimited expressions. You can prevent this behavior by explicitly delimiting the let or the if using parentheses or begin/end.