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Re: x = f(move(x))

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On 2011-09-08 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;
> }

I believe this should work because of the sequencing rules.

> I think it doesn't, because a move-constructor or move-assignment
> operator might modify its input parameter, but I'm not sure.

As of C++11 there has been added special wording to "5.17 Assignment
and compound assignment operators [expr.ass]":

"In all cases, the assignment is sequenced after the value
computation of the right and left operands, and before the value
computation of the assignment expression."

HTH & Greetings from Bremen,

Daniel Krügler


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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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