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Re: x = f(move(x))

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On 8 Sep., 23:47, Joe Gottman wrote:
>
> Suppose I have the following pair of functions:
>
>   string foo(const string &x); // Makes a copy of x, modifies, and returns it.
>   string foo(string &&x) // Modifies and returns x itself, since x is an rvalue.
>
> Given these functions, does the following code work?
>
>   int main()
>   {
>     string x = "hello";
>     x = foo(move(x));
>     return 0;
>   }

Yes. foo(move(x)) refers to some other temporary object. At this time,
x will be "empty" but it does not matter. x is still assignable.

> Of course, one way around this problem is to replace the line
>   x = foo(move(x))
> with the two lines
>   string temp = move(x);
>   x = f(move(temp));
>
> This would work, but it is slightly unwieldy, requiring a temporary
> variable and two calls to move.  It would be nice if there were a
> library function [...]

Let me stop you right there. As I said, there is no problem with your
1st code example. But I would like to suggest an alternative:

  string foo(string x) {  // no other overloads needed!
     std::reverse(x.begin(),x.end());
     return x;
  }
  :
  string z = ...;
  z = foo(move(z)); // x and foo(move(z)) will be move-constructed
automatically

The nice thing about move-enabled classes is that you don't need to
sprinkle rvalue references all over the place. x in foo and the
function's return value will automatically be move-constructed. Pass-
by-value should become more fashionable again in my opinion. Be sure
to check out Dave's "Want Speed? Pass by value!" article:
http://cpp-next.com/archive/2009/08/want-speed-pass-by-value/

Cheers!
SG


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x = f(move(x)) Joe Gottman <josephgottman@comcast.net> - 2011-09-08 14:47 -0700
  Re: x = f(move(x)) Daniel Krügler <daniel.kruegler@googlemail.com> - 2011-09-10 00:30 -0700
  Re: x = f(move(x)) SG <s.gesemann@gmail.com> - 2011-09-10 00:30 -0700

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