Can there be a difference between explicit operator= call and = operator?

Question

According to the C++ standard, can there be a difference between something.operator=(somethingElse) and something = somethingElse ?

The first appears in a template with T& something , and I'd like to know if it's safe to replace it with the more readable second version.

Answer 1

No there isn't. An explicit = operator invokes operator=() on a class that declares a suitable operator= method, exactly as if the operator= method was invoked directly.

This is true for all operators, not just = . That's the very definition of what an operator class method means: it specifies what gets executed when the corresponding operator is applied to an instance of the class.

Answer 2

Other than in copy-initialization contexts , the short operator expression is equivalent to the functional expression. Table 12 of the C++ standard draft describes the operator expressions:

Reproduced:

Subclause       Expression      As member function         As non-member function

[over.unary]  |     @a    |     (a).operator@ ( )      |    operator@(a)
[over.binary] |     a@b   |     (a).operator@ (b)      |    operator@(a, b)
[over.ass]    |     a=b   |     (a).operator= (b)      |
[over.sub]    |     a[b]  |     (a).operator[](b)      |
[over.ref]    |     a->   |     (a).operator->( )      |
[over.inc]    |     a@    |     (a).operator@ (0)      |    operator@(a, 0)

Where @ is a placeholder for the operator.

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Of cause, there are other contextual use of the = operator; in function declarations, such as defaulted member functions, deleted functions and = 0; // pure virtual = 0; // pure virtual

Answer 3

There are tiny differences, because a = b does not always invoke an explicit operator on object a :

• T a = b; . It is not an assignment but an initialization: it invokes a copy (or move) construction
• a = b; when a is an intrinsic object (integer, pointer, floating point, ...): the = operator is a builtin one.
• if T is trivially copyable, the assignement is allowed even when no operator = has been declared on the class, because the compiler use the default builtin assignement operator.

But if an operator = method exists and is selected by the overload resolution rules, the assignment operator will invoke it.