Calling assignment operator on an array

Question

I wonder how this code could work:

struct my_array
{
    int r[1000];
};

int main()
{
    my_array foo, bar;
    foo = bar;
}

Because the foo = bar call will invoke the constructor-provided operator= for the class, which will lazily apply it on every members. But arrays don't have an implementation for operator=, proof is, this code fails to compile:

int main()
{
    int a[1000], b[1000];
    a = b;
}

So how come my first code compiles?

Because structs, like classes, do have a default operator= that copies each member. — David Schwartz
– David Schwartz, Commented Dec 28, 2015 at 10:25
@DavidSchwartz yes? This is what I wrote, the default operator= is invoked, which will in turn apply operator= on every member, my question is: how does this work given arrays don't have operator= (i.e. can't be copied) — qdii
– qdii, Commented Dec 28, 2015 at 10:59
The default operator= copies each member. It doesn't necessarily call operator= on them unless that's the appropriate way to copy them. Why should a struct with int r[2]; copy differently from a struct with int r1, r2;? — David Schwartz
– David Schwartz, Commented Dec 28, 2015 at 11:00

juanchopanza · Accepted Answer · 2015-12-28 11:00:19Z

6

So how come my first code compiles?

The language specification says it has to work, and the compiler implements that behaviour.

The default assignment semantics are specified in clause 28 of § 12.8 [class.copy]. Specifically, the object's data members are assigned one by one. In the case of an array, this the array's elements are assigned one by one.

The implicitly-defined copy/move assignment operator for a non-union class X performs memberwise copy- /move assignment of its subobjects. ...

and

— if the subobject is an array, each element is assigned, in the manner appropriate to the element type;

(emphasis mine)

Note that in your particular example, the first code sample invokes undefined behaviour because the elements of bar are uninitialized at the moment you read from them here:

foo = bar; // UB: bar.r uninitialized

You can fix that by suitable initializing bar:

my_array foo;
my_array bar{};

edited Dec 28, 2015 at 11:00

answered Dec 28, 2015 at 10:25

juanchopanza

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2 Comments

qdii Over a year ago

So the standard specifies the meaning of "copying" an array. I guess the only reason why operator= isn't defined for an array is that it would cause a mess when arrays decay into pointers?

juanchopanza Over a year ago

@qdii It is a mess inherited from C. It would be hard to reconcile copy of arrays with pointer decay (and particularly array function parameter adjustment), although I don't see obvious problems with assignment. But it is the way it is and it can't be changed.

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