How Are C++ Array Members Handled in Copy Control Functions

How are C++ array members handled in copy control functions?

This is what the standard says in 12.8 (Copying class objects). Copy construction:

Each subobject is copied in the manner appropriate to its type:

• if the subobject is of class type, the copy constructor for the class is used;
• if the subobject is an array, each element is copied, in the manner appropriate to the element type;
• if the subobject is of scalar type, the built-in assignment operator is used.

Copy assignment:

Each subobject is assigned in the manner appropriate to its type:

• if the subobject is of class type, the copy assignment operator for the class is used (as if by explicit qualification; that is, ignoring any possible virtual overriding functions in more derived classes);
• if the subobject is an array, each element is assigned, in the manner appropriate to the element type;
• if the subobject is of scalar type, the built-in assignment operator is used.

c++ does implicit copy constructor copy array member variable?

Yes and yes is the answer. This is also true of structs in C.

typedef struct {
    int a[100];
} S;

S s1;
s1.a[0] = 42;
S s2;
s2 = s1;    // array copied

I thought array was non copyable

The defaulted copy constructor and assignment operator use copy construction and assignment individually on each member. When there's an array, copy construction or assignment is used on each element of the array (which is quite well-defined).

Here's the rule, from section 12.8 ([class.copy]):

The implicitly-defined copy/move constructor for a non-union class X performs a memberwise copy/move of its bases and members. [ Note: brace-or-equal-initializers of non-static data members are ignored. See also the example in 12.6.2. — end note ] The order of initialization is the same as the order of initialization of bases and members in a user-defined constructor (see 12.6.2). Let x be either the parameter of the
constructor or, for the move constructor, an xvalue referring to the parameter. Each base or non-static data member is copied/moved in the manner appropriate to its type:

• if the member is an array, each element is direct-initialized with the corresponding subobject of x;
• if a member m has rvalue reference type T&&, it is direct-initialized with static_cast<T&&>(x.m);
• otherwise, the base or member is direct-initialized with the corresponding base or member of x.

and

The implicitly-defined copy/move assignment operator for a non-union class X performs memberwise copy/move assignment of its subobjects. The direct base classes of X are assigned first, in the order of their declaration in the base-specifier-list, and then the immediate non-static data members of X are assigned, in the order in which they were declared in the class definition. Let x be either the parameter of the function or, for the move operator, an xvalue referring to the parameter. Each subobject is assigned in the manner appropriate to its type:

• if the subobject is of class type, as if by a call to operator= with the subobject as the object expression and the corresponding subobject of x as a single function argument (as if by explicit qualification; that is, ignoring any possible virtual overriding functions in more derived classes);
• if the subobject is an array, each element is assigned, in the manner appropriate to the element type;
• if the subobject is of scalar type, the built-in assignment operator is used.

The rule for signature selection between C::C(const C&) vs C::C(C&) et al also includes language referring to the array element type.

Are synthesized copy control members always public?

Yes, you can get the synthesized special functions, but with different access controls:

class Foo {
  private:
     Foo() = default;
};

Since it's still declared private, it's not public, but you don't need to define it.

c++: how does memberwise assignment work?

A b=a;

Is not assignment, it is called as Copy Initialization.

The implicitly generated copy constructor is called to create an new object b from the existing object a.

The implicitly generated copy constructor makes a copy of the array member.

For completeness I am going to add here the standard citation from the marked duplicate.

C++03 Standard: 12.8 (Copying class objects)

Each subobject is copied in the manner appropriate to its type:

• if the subobject is of class type, the copy constructor for the class is used;
• if the subobject is an array, each element is copied, in the manner appropriate to the element type;
• if the subobject is of scalar type, the built-in assignment operator is used.

How is memory allocated to instances of class with array member?

All data members of a class have the same storage-duration as the class' object created. Hence in:

int main(){
   Arr temp; 
   doSomething(temp);
}

The data member temp.arr which is of type int[10]; also has automatic storage-duration.. (or what is inappropriately known as "stack object")

---

Arr doSomething(Arr &arg){
   return arg; //copy of arg created; 'copy=new int[10]' at this point?
}

No, there is no heap allocation here. rather a member-wise copy by the implicit copy constructor. See How are C++ array members handled in copy control functions?

When is an array name or a function name 'converted' into a pointer ? (in C)

An expression of array type is implicitly converted to a pointer to the array object's first element unless it is:

• The operand of the unary & operator;
• The operand of sizeof; or
• A string literal in an initializer used to initialize an array object.

An examples of the third case are:

char arr[6] = "hello";

"hello" is an array expression, of type char[6] (5 plus 1 for the '\0' terminator). It's not converted to an address; the full 6-byte value of of "hello" is copied into the array object arr.

On the other hand, in this:

char *ptr = "hello";

the array expression "hello" "decays" to a pointer to the 'h', and that pointer value is used to initialize the pointer object ptr. (It should really be const char *ptr, but that's a side issue.)

An expression of function type (such as a function name) is implicitly converted to a pointer to the function unless it is:

• The operand of the unary & operator; or
• The operand of sizeof (sizeof function_name is illegal, not the size of a pointer).

That's it.

In both cases, no pointer object is created. The expression is converted to ("decays" to) a pointer value, also known as an address.

(The "conversion" in both these cases isn't an ordinary type conversion like the one specified by a cast operator. It doesn't take the value of an operand and use it to compute the value of the result, as would happen for an int-to-float conversion. Rather an expression of array or function type is "converted" at compile time to an expression of pointer type. In my opinion the word "adjusted" would have been clearer than "converted".)

Note that both the array indexing operator [] and the function call "operator" () require a pointer. In an ordinary function call like func(42), the function name func "decays" to a pointer-to-function, which is then used in the call. (This conversion needn't actually be performed in the generated code, as long as the function call does the right thing.)

The rule for functions has some odd consequences. The expression func is, in most contexts, converted to a pointer to the function func. In &func, func is not converted to a pointer, but & yields the function's address, i.e., a pointer value. In *func, func is implicitly converted to a pointer, then * dereferences it to yield the function itself, which is then (in most contexts) converted to a pointer. In ****func, this happens repeatedly.

(A draft of the C11 standard says that there's another exception for arrays, namely when the array is the operand of the new _Alignof operator. This is an error in the draft, corrected in the final published C11 standard; _Alignof can only be applied to a parenthesized type name, not to an expression.)

The address of an array and the address of its first member:

int arr[10];
&arr;    /* address of entire array */
&arr[0]; /* address of first element */

are the same memory address, but they're of different types. The former is the address of the entire array object, and is of type int(*)[10] (pointer to array of 10 ints); the latter is of type int*. The two types are not compatible (you can't legally assign an int* value to an int(*)[10] object, for example), and pointer arithmetic behaves differently on them.

There's a separate rule that says that a declared function parameter of array or function type is adjusted at compile time (not converted) to a pointer parameter. For example:

void func(int arr[]);

is exactly equivalent to

void func(int *arr);

These rules (conversion of array expressions and adjustment of array parameters) combine to create a great deal of confusion regarding the relationship between arrays and pointers in C.

Section 6 of the comp.lang.c FAQ does an excellent job of explaining the details.

The definitive source for this is the ISO C standard. N1570 (1.6 MB PDF) is the latest draft of the 2011 standard; these conversions are specified in section 6.3.2.1, paragraphs 3 (arrays) and 4 (functions). That draft has the erroneous reference to _Alignof, which doesn't actually apply.

Incidentally, the printf calls in your example are strictly incorrect:

int fruits[10];
printf("Address IN constant pointer is %p\n",fruits);
printf("Address OF constant pointer is %p\n",&fruits); 

The %p format requires an argument of type void*. If pointers of type int* and int(*)[10] have the same representation as void* and are passed as arguments in the same way, as is the case for most implementations, it's likely to work, but it's not guaranteed. You should explicitly convert the pointers to void*:

int fruits[10];
printf("Address IN constant pointer is %p\n", (void*)fruits);
printf("Address OF constant pointer is %p\n", (void*)&fruits);

So why is it done this way? The problem is that arrays are in a sense second-class citizens in C. You can't pass an array by value as an argument in a function call, and you can't return it as a function result. For arrays to be useful, you need to be able to operate on arrays of different lengths. Separate strlen functions for char[1], for char[2], for char[3], and so forth (all of which are distinct types) would be impossibly unwieldy. So instead arrays are accessed and manipulated via pointers to their elements, with pointer arithmetic providing a way to traverse those elements.

If an array expression didn't decay to a pointer (in most contexts), then there wouldn't be much you could do with the result. And C was derived from earlier languages (BCPL and B) that didn't necessarily even distinguish between arrays and pointers.

Other languages are able to deal with arrays as first-class types but doing so requires extra features that wouldn't be "in the spirit of C", which continues to be a relatively low-level language.

I'm less sure about the rationale for treating functions this way. It's true that there are no values of function type, but the language could have required a function (rather than a pointer-to-function) as the prefix in a function call, requiring an explicit * operator for an indirect call: (*funcptr)(arg). Being able to omit the * is a convenience, but not a tremendous one. It's probably a combination of historical inertia and consistency with the treatment of arrays.

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