指定表达式右侧的数组类型

Question

With the following:

typedef struct Person_ {
    char* name;
    char* parents[2];
} Person;

int main(void)
{
    Person jack = {.name="Jack", .parents={"Jim", "Julia"}};
    Person tom = {.name="Tom", .parents={"Terry", "Tina"}};
    Person friends[2] = (Person[2]) {jack, tom};
}

I get this error:

error: array initialized from non-constant array expression
Person friends[2] = (Person[2]) {jack, tom};

I know that I can initialize friends as:

Person friends[2] = {jack, tom};

But why isn't it possible to annotate the type on the right side with an array like I would be able to (though redundantly) with another type such as:

int x = (int) 4;
char *y = (char*) "Hi";

// invalid
// int z[2] = (int[2]) {1,2}; 

Answer 1

In this statement

Person friends[2] = (Person[2]) {jack, tom};

(Person[2]) {jack, tom} is a compound literal.

From C11 Standard - Compound literals#6.5.2.5p5

5 The value of the compound literal is that of an unnamed object initialized by the initializer list. ....

So, this (Person[2]) {jack, tom} will result in an unnamed array of Person type initialised from the initialiser list {jack, tom} . You cannot intialize an array from another array.

From C11 Standard - Compound literals#6.7.9p16

16 Otherwise, the initializer for an object that has aggregate or union type shall be a brace- enclosed list of initializers for the elements or named members.

This {jack, tom} is initialiser list whereas this (Person[2]) {jack, tom} is compound literal.

The compound literal results in an unnamed array, initialises it with given values and creates a pointer point to first element of array. So, in this statement

int *a1  = (int[2]) {1,2};

(int[2]) {1,2} will result in an unnamed array of 2 int initialised with value {1, 2} and create pointer to initial element of unnamed array.

From C11 Standard#6.3.2.1p3

3 Except when it is the operand of the sizeof operator, the _Alignof operator, or the unary & operator, or is a string literal used to initialize an array, an expression that has type ''array of type'' is converted to an expression with type ''pointer to type'' that points to the initial element of the array object and is not an lvalue. ....