boost lambda for_each / transform puzzle

Question

Does anybody know why

  vector<int> test(10);
  int a=0;

  for_each(test.begin(),test.end(),(_1+=var(a),++var(a)));

  for_each(test.begin(),test.end(),(cout << _1 << " "));
  cout << "\n"

Gives : "0 1 2 3 4 5 6 7 8 9"

but

  transform(test.begin(),test.end(),test.begin(), (_1+=var(a),++var(a)));
  ...(as before)

Gives : "1 2 3 4 5 6 7 8 9 10"

?

Answer 1

Comma operator evaluates left to right, so the result of the

_1+=var(a), ++var(a)

is ++var(a), which you'll store using the transform version.

• for_each:

  _1 += var(a) is evaluated, updating your sequence (via the lambda _1), then ++var(a) is evaluated, but this has no effect on your sequence.

• transform:

  _1+=var(a) is evaluated, updating your sequence (just like before), then ++var(a) is evaluated, this also gives the result of the whole expression, then that is used to update your sequence again (via the transform)

Answer 2

Essentially, in the for_each you provide a function with a side-effect, while in the transform, your use the returnvalue of a function.

In your case, you reuse the same function. Since operator += happens to have a return value, this is the one used as a result of the transformation.