In fact, it works in exactly the same way against integer and string.

Does not. Try this:

$\ = "\n";
print ~3;
print ~"3";
[download]

Result:

4294967292
Ì

Now, if it worked exactly the same way for integers and for strings, how come the intermediate result is different?

I stand by my point: even though they work differently (as shown here), the net result of the double not is exactly the same for both: the original value. That means that an integer produces the original integer, and that a string produces the original string.

Perhaps you mean something other than I do?

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