I've seen a statement according to which
string noun("ants");
string noun = "ants";
are exactly equivalent.
This is contrary to my intuition: I thought in the second case a conversion occurs (via a constructor of the string class), then the assignment operator gets called with an argument of class string. What is there actually happening ?
For initialization of a variable definition using assignment, like
std::string a = "foo";
Here two objects are created: The variable you define (a in my example) and a temporary object (for the string "foo"). Then the copy-constructor of your variable (a) is called with the temporary object, and then the temporary object is destructed. The copy-assignment operator is not called.
However, the copying can be avoided if the compiler uses copy elision, which is an optimization technique to avoid unnecessary copies and copying. The copy-constructor needs to be there anyway, even if it's not called.
For the definition
std::string a("foo");
there is a better constructor which takes a pointer to a constant string, which the literal "foo" can be seen as (string literals are actually constant arrays of char, but like all arrays they decay to pointers).
deleteing the copy-constructor or making it private, then your compiler should give an error for the second version.I thought it would be harder to find, but here goes:
§12.6.1 Explicit initialization ([class.expl.init])
complex a(1); // initialize by a call of
// complex(double)
complex b = a; // initialize by a copy of a
As @pmr pointed out, initialization is not assignment, thus no call to an assignment operator is made. Refer to the rest of 12.6.1 for details.
Valuable information can also be found in §8.5 ([dcl.init]), with 8.5.1 denoting
initializer:
brace-or-equal-initializer
( expression-list )
brace-or-equal-initializer:
= initializer-clause
braced-init-list
and 8.5.2 showing examples of initializers.
Disclaimer: I only had n3485 laying around, but I doubt it changed in C++14
op=is made. Compare tostring noun; noun = "ants";