Inheritance: Constructor, initialization C, similar to array members of the foundation class in C 11

struct Base //in my real scenario Base class can not be changed
 {
 int a;
 double b[10];
};

struct Child: Base
{
 Child(int aa, double bb[10]): Base{aa} {} //This works fine
 Child(int aa, double bb[10]): Base{aa, bb} {} //this is not working
};

In Child’s constructor, bb is not an array: because decay, it is just a pointer. And you cannot initialize an array with a pointer (b is in Base).

In two classes Using std::array instead of the original array will solve your problem:

struct Base
{
 int a; 
 std::array b;
};

struct Child: Base
{
 Child(int aa, std:: array bb): Base(aa, bb) {}
};

However, since you mentioned that Base cannot be modified, you must manually copy the elements (in general , You can also use move, but the basic type is meaningless):

#include 
#in clude 
#include 

struct Base {
 int a;
 double b[10];
};

struct Child: Base {
 Child(int aa, std::array bb): Base{aa} {
 std::copy(bb.begin() , bb.end(), b);
 }
};

int main() {
 auto child = Child(3, {2, 3, 4});

 for (auto it: child.b) {
 std::cout << it << "";
 }
} 
Live on Coliru
 
You can also use references instead of std::array to achieve, but the syntax is a bit complicated:
 
 
Child( int aa, double const (&bb)[10]): Base{aa} {
 std::copy(&bb[0], &bb[10], b);
}
 < /div>
Consider the following code: 

struct Base //in my real scenario Base class can not be changed
{ 
 int a;
 double b[10];
};

struct Child: Base
{
 Child(int aa, double bb[10]): Base{aa} {} //This works fine
 Child(int aa, double bb[10]): Base{aa, bb} {} //this is not working
};
 The child's second constructor does not work. I get the error "Array must be initialized with a bracket-enclosed initializer".
How to initialize b in Child without changing the Base class (e.g. using vector instead of c-like array, I don’t allow it)

In Child’s constructor, bb is not an array: because decay, it’s just a pointer. And you You cannot initialize an array with a pointer (b is in Base). 
Using std::array in both classes instead of the original array will solve your problem: 

struct Base
{
 int a;
 std::array b;
};< br />
struct Child: Base
{
 Child(int aa, std::array bb): Base{aa, bb} {}
} ;
However, since you mentioned that Base cannot be modified, you must manually copy the elements (in general, you can also use move, but the basic type does not make sense):
< /p> 
#include 
#include 
#include 

struct Base {
 int a;
 double b[10];
};

struct Child: Base {
 Child(int aa, std::array bb): Base{aa } {
 std::copy(bb.begin(), bb.end(), b);
 }
};

int main() { 
 auto child = Child(3, {2, 3, 4});

 for (auto it: child.b) {
 std::cout << it < <"";
 }
}< /pre> 
Live on Coliru
 
You can also use references instead of std::array to achieve, but the syntax is a bit complicated:
 
 
Child (int aa, double const (&bb)[10]): Base{aa} {
 std::copy(&bb[0], &bb[10], b);
}