Consider the following code:
#include <array>
struct A
{
int a;
int b;
};
static std::array<A, 4> x1 =
{
{ 1, 2 },
{ 3, 4 },
{ 5, 6 },
{ 7, 8 }
};
static std::array<A, 4> x2 =
{
{
{ 1, 2 },
{ 3, 4 },
{ 5, 6 },
{ 7, 8 }
}
};
static std::array<A, 4> x3 =
{
A{ 1, 2 },
A{ 3, 4 },
A{ 5, 6 },
A{ 7, 8 }
};
static std::array<A, 4> x4 =
{
A{ 1, 2 },
{ 3, 4 },
{ 5, 6 },
{ 7, 8 }
};
Compiling with gcc:
$ gcc -c --std=c++11 array.cpp
array.cpp:15:1: error: too many initializers for ‘std::array<A, 4ul>’
};
^
$
NB1: Commenting out the first initialisation statement, the code compiles without error.
NB2: Converting all the initialisation to constructor calls yields the same results.
NB3: MSVC2015 behaves the same.
I can see why the first initialisation fails to compile, and why the second and third are OK. (e.g. See C++11: Correct std::array initialization?.)
My question is: Why exactly does the final initialisation compile?
Short version: An initializer-clause that starts with { stops brace-elision. This is the case in the first example with {1,2}, but not in the third nor fourth which use A{1,2}. Brace-elision consumes the next N initializer-clauses (where N is dependent on the aggregate to be initialized), which is why only the first initializer-clause of the N must not begin with {.
In all implementations of the C++ Standard Library I know of, std::array is a struct which contains a C-style array. That is, you have an aggregate which contains a sub-aggregate, much like
template<typename T, std::size_t N>
struct array
{
T __arr[N]; // don't access this directly!
};
When initializing a std::array from a braced-init-list, you'll therefore have to initialize the members of the contained array. Therefore, on those implementations, the explicit form is:
std::array<A, 4> x = {{ {1,2}, {3,4}, {5,6}, {7,8} }};
The outermost set of braces refers to the std::array struct; the second set of braces refers to the nested C-style array.
C++ allows in aggregate initialization to omit certain braces when initializing nested aggregates. For example:
struct outer {
struct inner {
int i;
};
inner x;
};
outer e = { { 42 } }; // explicit braces
outer o = { 42 }; // with brace-elision
The rules are as follows (using a post-N4527 draft, which is post-C++14, but C++11 contained a defect related to this anyway):
Braces can be elided in an initializer-list as follows. If the initializer-list begins with a left brace, then the succeeding comma-separated list of initializer-clauses initializes the members of a subaggregate; it is erroneous for there to be more initializer-clauses than members. If, however, the initializer-list for a subaggregate does not begin with a left brace, then only enough initializer-clauses from the list are taken to initialize the members of the subaggregate; any remaining initializer-clauses are left to initialize the next member of the aggregate of which the current subaggregate is a member.
Applying this to the first std::array-example:
static std::array<A, 4> x1 =
{
{ 1, 2 },
{ 3, 4 },
{ 5, 6 },
{ 7, 8 }
};
This is interpreted as follows:
static std::array<A, 4> x1 =
{ // x1 {
{ // __arr {
1, // __arr[0]
2 // __arr[1]
// __arr[2] = {}
// __arr[3] = {}
} // }
{3,4}, // ??
{5,6}, // ??
...
}; // }
The first { is taken as the initializer of the std::array struct. The initializer-clauses {1,2}, {3,4} etc. then are taken as the initializers of the subaggregates of std::array. Note that std::array only has a single subaggregate __arr. Since the first initializer-clause {1,2} begins with a {, the brace-elision exception does not occur, and the compiler tries to initialize the nested A __arr[4] array with {1,2}. The remaining initializer-clauses {3,4}, {5,6} etc. do not refer to any subaggregate of std::array and are therefore illegal.
In the third and fourth example, the first initializer-clause for the subaggregate of std::array does not begin with a {, therefore the brace elision exception is applied:
static std::array<A, 4> x4 =
{
A{ 1, 2 }, // does not begin with {
{ 3, 4 },
{ 5, 6 },
{ 7, 8 }
};
So it is interpreted as follows:
static std::array<A, 4> x4 =
{ // x4 {
// __arr { -- brace elided
A{ 1, 2 }, // __arr[0]
{ 3, 4 }, // __arr[1]
{ 5, 6 }, // __arr[2]
{ 7, 8 } // __arr[3]
// } -- brace elided
}; // }
Hence, the A{1,2} causes all four initializer-clauses to be consumed to initialize the nested C-style array. If you add another initializer:
static std::array<A, 4> x4 =
{
A{ 1, 2 }, // does not begin with {
{ 3, 4 },
{ 5, 6 },
{ 7, 8 },
X
};
then this X would be used to initialize the next subaggregate of std::array. E.g.
struct outer {
struct inner {
int a;
int b;
};
inner i;
int c;
};
outer o =
{ // o {
// i {
1, // a
2, // b
// }
3 // c
}; // }
Brace-elision consumes the next N initializer-clauses, where N is defined via the number of initializers required for the (sub)aggregate to be initialized. Therefore, it only matters whether or not the first of those N initializer-clauses starts with a {.
More similarly to the OP:
struct inner {
int a;
int b;
};
struct outer {
struct middle {
inner i;
};
middle m;
int c;
};
outer o =
{ // o {
// m {
inner{1,2}, // i
// }
3 // c
}; // }
Note that brace-elision applies recursively; we can even write the confusing
outer o =
{ // o {
// m {
// i {
1, // a
2, // b
// }
// }
3 // c
}; // }
Where we omit both the braces for o.m and o.m.i. The first two initializer-clauses are consumed to initialize o.m.i, the remaining one initializes o.c. Once we insert a pair of braces around 1,2, it is interpreted as the pair of braces corresponding to o.m:
outer o =
{ // o {
{ // m {
// i {
1, // a
2, // b
// }
} // }
3 // c
}; // }
Here, the initializer for o.m does start with a {, hence brace-elision does not apply. The initializer for o.m.i is 1, which does not start with a {, hence brace-elision is applied for o.m.i and the two initializers 1 and 2 are consumed.
来源：https://stackoverflow.com/questions/31449898/initialisation-of-stdarray