C++ entangled shared pointer

Question

I have found code below in "The C++ programming language, 4th edition", chapter 17.5.1.3

struct S2 {
    shared_ptr<int> p;
};

S2 x {new int{0}};

void f()
{
    S2 y {x};                // ‘‘copy’’ x
    ∗y.p = 1;                // change y, affects x
    ∗x.p = 2;                // change x; affects y
    y.p.reset(new int{3});   // change y; affects x
    ∗x.p = 4;                // change x; affects y
}

I don't understand the last comment, indeed y.p should point to a new memory address after the reset() call, and so

    ∗x.p = 4;

should let y.p unchanged, isn't it?

Thanks

clang seems to disagree [it says y.p is 3 at the end](http://coliru.stacked-crooked.com/a/c687c657a22cb323) — Borgleader, Dec 11 '14 at 22:06
It is large book there is bound to be erroneous examples, see my recent [question here for another example](http://stackoverflow.com/q/27158812/1708801) as Bjarne says [here](http://www.stroustrup.com/4th.html) *Feel free to send me suggested errata. I will fix every mistake that can be fixed within the constrains of the book*. How to contact him can be found on the [main page](http://www.stroustrup.com/). — Shafik Yaghmour, Dec 11 '14 at 22:15

Bill Lynch · Accepted Answer · 2014-12-11T22:14:11.880

The book is wrong and you are correct. You might consider sending this to Bjarne so it could be fixed in the next printing.

The correct comments might be:

S2 y {x};                // x.p and y.p point to the same int.
*y.p = 1;                // changes the value of both *x.p and *y.p
*x.p = 2;                // changes the value of both *x.p and *y.p
y.p.reset(new int{3});   // x.p and y.p point to different ints.
*x.p = 4;                // changes the value of only *x.p

C++ entangled shared pointer

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