Extract Method creates an unnecessary parameter for variable

[DanTup]

This was originally raised at Dart-Code/Dart-Code#5373 by @stephane-archer.

If you extract code that sets a variable, it becomes a return value for the extracted method. The original variable is also passed as an argument, even if it has never been read in the extracted code and could just be declared inside the new function.

For example:

void f() {
  int? i;
  i = 1; // extract this line
  print(i);
}

Becomes:

void f() {
  int? i;
  i = g(i);
  print(i);
}

int? g(int? i) {
  i = 1;
  return i;
}

But since i is not read by the extracted method (only written), the parameter could be dropped and instead the variable just declared inside the extracted function:

void f() {
  int? i;
  i = g();
  print(i);
}

int? g() {
  int? i = 1;
  return i;
}

[bwilkerson]

The request is valid, but I suspect that it doesn't effect that many users. If I'm wrong, please upvote the issue as a signal.

[stephane-archer]

I think the code we extract can be much complex and this will create strange extracted code results.

Users might not recognize this pattern by looking at this simple example but they would see a lot of extra added parameters when they extract code.

I suspect this affects a lot of users but they just don't expect the code extractor tool to be smart and are okay with rewriting this after the tool has done a first step

[lrhn]

(Independently of that parameter, it would be nice if the return type of the function was int, since that is the type of i after the expression has been evaluated.
Otherwise:

int? i;
i = 1;
print(i.toRadixString(16));

won't be able to convert the i = 1 to

 ...
 i = g(i);
...

int g(int? i) {
  i = 1;
  return i;
}

But that'd be another issue, so this is just a drive-by comment.)