The difference between *s.pd and s->pd

Question

I was reading about copy constructors for structs and i found this example:

#include <iostream> #include <string> using namespace std; struct SomeData { int * pd; string id; SomeData(SomeData & ref) { cout << "Copy Constructor called" << endl; pd = new int (*ref.pd); id = "Copy Constructed"; } SomeData(string name) { pd = new int(0); id = name; cout << "Constructor for " << id << endl; }; ~SomeData() { cout << "Destructor for " << id << endl; delete pd; } }; int main() { SomeData s("First"); *s.pd = 9; SomeData s2=s; cout << *s2.pd << endl; return 0; }

in the main, the member pd of SomeData is accessed using the dereference, but why is that, isn't the correct way is

s->pd=9;

why was it written like that in the example?

Take a look at the rule-of-3 and the rule-of-zero. The implicitly defined op= is catastrophically wrong. — Deduplicator
– Deduplicator, Commented Dec 25, 2014 at 3:09

Jules · Accepted Answer · 2014-12-24 20:04:30Z

This is an example of operator precedence not giving quite the results you expect. The example has the code

*s.pd = 9;

which is equivalent to the following:

*(s.pd) = 9;

You query why it isn't written as

s->pd = 9;

but this is equivalent to

(*s).pd = 9;

That is, the code in the example dereferences the value of pd that is a member of s, whereas the code you are asking about dereferences s (which is not legal as s is not a pointer and doesn't have an overload for the "->" operator).

This is because ".", the member selection operator binds more strongly than "*", the dereference operator.

See http://en.cppreference.com/w/cpp/language/operator_precedence for a complete list; "." has a precedence of 2, while "*" has a precedence of 3. Only "::" binds more strongly than ".".

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