copy constructor and operator = in c++ - c++

Question

I have a question about the following code:
let's say I have a class P that has a copy constructor and a regular constructor the receives one string value.
I have the following code:
P doSomething(){
P p("myValue");
return p;
}
int main(){
P m=doSomething();
return 1;
}
why isn't copy constructor invoked at the return p of the doSomething() function?
the call P m=doSomething() - does it suppose to call the copy constructor or the operator=?
in case it's operator =, what is the difference of this code and the following:
P new_val=("newVal");
p m=new_val;
(i know here the call is for copy constructor)
Thanks,
Mary

Answer 1

why isn't copy constructor invoked at the return p of the doSomething() function?
The standard allows this copy to be elided. Google for [N]RVO. On certain compilers this happens only when optimizing, on others it is part of the calling conventions and thus happens always.
the call P m=doSomething() - does it suppose to call the copy constructor or the operator=?
T t = x; is "syntactic sugar" (in the sense that the T(x) is happening implicitly) for T t(T(x)) and thus has -- despite the = -- nothing to do with operator=. Note that also here the additional temporary may be elided, thus no copy ctor is called.
in case it's operator =, what is the difference of this code and the following:
This code makes no sense, what did you really mean?
P new=("newVal");
p m=new;

Answer 2

I made a small sample for demonstration.
void print(char* text, int ident)
{
for(int i = 0 ; i < ident ; i++)
{
printf(" ");
} // end for
fprintf(stdout, text);
fprintf(stdout, "\n");
fflush(stdout);
}
class P
{
public:
char* _str;
P (P& arg)
{
print("copy constructor", 2);
arg._str = this->_str;
}
P (char* str)
{
print("constructor", 2);
_str = str;
}
};
P doSomething(){
print("do something - function", 1);
P p("myValue");
print("before return p", 1);
return p;
}
int main(int argc, char* argv[])
{
print("start", 0);
P m=doSomething();
print("stop - call return", 0);
return 1;
}
this returns
start
do something - function
constructor
before return p
copy constructor
stop - call return
so copy constructor WILL BE CALLED

Answer 3

1) why isn't copy constructor invoked at the "return p" of the doSomething() function?
it should - try printing something to the console in the copy c-tor to see if it is really running.
2) the call P m=doSomething() - does it suppose to call the copy constructor or the operator=?
should call the operator=. again, use the debug message print from within the method to check
3) in case it's operator =, what is the difference of this code and the following:
P new=("newVal"); p m=new; (i know here the call is for copy constructor)
did you miss something on the code snippet? i think it won't compile