CODEWITHSUNDEEP
c++arraysassignment-operator
qdii
qdii

Reputation: 12963

Calling assignment operator on an array

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?

Upvotes: 1

Views: 1647

Answers (1)

juanchopanza
juanchopanza

Reputation: 227390

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{};

Upvotes: 6

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