为什么 C 和 C + + 支持在结构中按成员分配数组，而不是一般情况下？

我知道不支持数组的成员分配，因此下面的方法不起作用:

int num1[3] = {1,2,3};
int num2[3];
num2 = num1; // "error: invalid array assignment"

我只是接受了这个事实，认为该语言的目的是提供一个开放式框架，并让用户决定如何实现某些东西，比如数组的复制。

不过，以下措施确实有效:

struct myStruct { int num[3]; };
struct myStruct struct1 = {{1,2,3}};
struct myStruct struct2;
struct2 = struct1;

数组 num[3]以成员方式从它在 struct1中的实例分配到它在 struct2中的实例中。

为什么结构支持按成员分配数组，但通常不支持？

编辑 : 线程 Struct 中的字符串-复制/赋值问题？中 Roger Pate的注释似乎指向了答案的大致方向，但我自己并不知道该如何确认。

编辑2 : 许多优秀的回答。我选择 Luther Blissett是因为我主要想知道这种行为背后的哲学或历史原理，但是 James McNellis对相关规范文档的参考也很有用。

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小开

In this link: http://www2.research.att.com/~bs/bs_faq2.html there's a section on array assignment:

The two fundamental problems with arrays are that

  • an array doesn't know its own size
  • the name of an array converts to a pointer to its first element at the slightest provocation

And I think this is the fundamental difference between arrays and structs. An array variable is a low level data element with limited self knowledge. Fundamentally, its a chunk of memory and a way to index into it.

So, the compiler can't tell the difference between int a[10] and int b[20].

Structs, however, do not have the same ambiguity.

小开

Concerning the assignment operators, the C++ standard says the following (C++03 §5.17/1):

There are several assignment operators... all require a modifiable lvalue as their left operand

An array is not a modifiable lvalue.

However, assignment to a class type object is defined specially (§5.17/4):

Assignment to objects of a class is defined by the copy assignment operator.

So, we look to see what the implicitly-declared copy assignment operator for a class does (§12.8/13):

The implicitly-defined copy assignment operator for class X performs memberwise assignment of its subobjects. ... Each subobject is assigned in the manner appropriate to its type:
...
-- if the subobject is an array, each element is assigned, in the manner appropriate to the element type
...

So, for a class type object, arrays are copied correctly. Note that if you provide a user-declared copy assignment operator, you cannot take advantage of this, and you'll have to copy the array element-by-element.

The reasoning is similar in C (C99 §6.5.16/2):

An assignment operator shall have a modifiable lvalue as its left operand.

And §6.3.2.1/1:

A modifiable lvalue is an lvalue that does not have array type... [other constraints follow]

In C, assignment is much simpler than in C++ (§6.5.16.1/2):

In simple assignment (=), the value of the right operand is converted to the type of the assignment expression and replaces the value stored in the object designated by the left operand.

For assignment of struct-type objects, the left and right operands must have the same type, so the value of the right operand is simply copied into the left operand.

小开
最佳答案

Here's my take on it:

The Development of the C Language offers some insight in the evolution of the array type in C:

I'll try to outline the array thing:

C's forerunners B and BCPL had no distinct array type, a declaration like: 

auto V[10] (B)
or
let V = vec 10 (BCPL)

would declare V to be a (untyped) pointer which is initialized to point to an unused region of 10 "words" of memory. B already used * for pointer dereferencing and had the [] short hand notation, *(V+i) meant V[i], just as in C/C++ today. However, V is not an array, it is still a pointer which has to point to some memory. This caused trouble when Dennis Ritchie tried to extend B with struct types. He wanted arrays to be part of the structs, like in C today: 

struct {
int inumber;
char name[14];
};

But with the B,BCPL concept of arrays as pointers, this would have required the name field to contain a pointer which had to be initialized at runtime to a memory region of 14 bytes within the struct. The initialization/layout problem was eventually solved by giving arrays a special treatment: The compiler would track the location of arrays in structures, on the stack etc. without actually requiring the pointer to the data to materialize, except in expressions which involve the arrays. This treatment allowed almost all B code to still run and is the source of the "arrays convert to pointer if you look at them" rule. It is a compatiblity hack, which turned out to be very handy, because it allowed arrays of open size etc.

And here's my guess why array can't be assigned: Since arrays were pointers in B, you could simply write: 

auto V[10];
V=V+5;

to rebase an "array". This was now meaningless, because the base of an array variable was not a lvalue anymore. So this assigment was disallowed, which helped to catch the few programs that did this rebasing on declared arrays. And then this notion stuck: As arrays were never designed to be first class citized of the C type system, they were mostly treated as special beasts which become pointer if you use them. And from a certain point of view (which ignores that C-arrays are a botched hack), disallowing array assignment still makes some sense: An open array or an array function parameter is treated as a pointer without size information. The compiler doesn't have the information to generate an array assignment for them and the pointer assignment was required for compatibility reasons. Introducing array assignment for the declared arrays would have introduced bugs though spurious assigments (is a=b a pointer assignment or an elementwise copy?) and other trouble (how do you pass an array by value?) without actually solving a problem - just make everything explicit with memcpy! 

/* Example how array assignment void make things even weirder in C/C++,
if we don't want to break existing code.
It's actually better to leave things as they are...
*/
typedef int vec[3];


void f(vec a, vec b)
{
vec x,y;
a=b; // pointer assignment
x=y; // NEW! element-wise assignment
a=x; // pointer assignment
x=a; // NEW! element-wise assignment
}

This didn't change when a revision of C in 1978 added struct assignment ( http://cm.bell-labs.com/cm/cs/who/dmr/cchanges.pdf ). Even though records were distinct types in C, it was not possible to assign them in early K&R C. You had to copy them member-wise with memcpy and you could pass only pointers to them as function parameters. Assigment (and parameter passing) was now simply defined as the memcpy of the struct's raw memory and since this couldn't break exsisting code it was readily adpoted. As a unintended side effect, this implicitly introduced some kind of array assignment, but this happended somewhere inside a structure, so this couldn't really introduce problems with the way arrays were used.

小开

I know, everyone who answered are experts in C/C++. But I thought, this is the primary reason.

num2 = num1;

Here you are trying to change the base address of the array, which is not permissible.

and of course, struct2 = struct1;

Here, object struct1 is assigned to another object.

小开

Another reason no further efforts were made to beef up arrays in C is probably that array assignment would not be that useful. Even though it can be easily achieved in C by wrapping it in a struct (and the struct's address can be simply cast to the array's address or even the array's first element's address for further processing) this feature is rarely used. One reason is that arrays of different sizes are incompatible which limits the benefits of assignment or, related, passing to functions by value.

Most functions with array parameters in languages where arrays are first-class types are written for arrays of arbitrary size. The function then usually iterates over the given number of elements, an information that the array provides. (In C the idiom is, of course, to pass a pointer and a separate element count.) A function which accepts an array of just one specific size is not needed as often, so not much is missed. (This changes when you can leave it to the compiler to generate a separate function for any occurring array size, as with C++ templates; this is the reason why std::array is useful.)


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