zero-initializing elements of a std::array with default member initializer?
Suppose I have a class template like this:
template<typename T, size_t N>
struct S {
std::array<T,N> a;
};
Is there a default member initializer I can place on a:
template<typename T, size_t N>
struct S {
std::array<T,N> a = ???;
};
such that no matter what T is, the elements of a will always be initialized (never have indeterminant value)? ie Even if T is a primitive type like int.
c++ c++17
asked 5 hours ago
Andrew Tomazos
34.6k25132227
add a comment |
Suppose I have a class template like this:
template<typename T, size_t N>
struct S {
std::array<T,N> a;
};
Is there a default member initializer I can place on a:
template<typename T, size_t N>
struct S {
std::array<T,N> a = ???;
};
such that no matter what T is, the elements of a will always be initialized (never have indeterminant value)? ie Even if T is a primitive type like int.
c++ c++17
asked 5 hours ago
Andrew Tomazos
34.6k25132227
add a comment |
Suppose I have a class template like this:
template<typename T, size_t N>
struct S {
std::array<T,N> a;
};
Is there a default member initializer I can place on a:
template<typename T, size_t N>
struct S {
std::array<T,N> a = ???;
};
such that no matter what T is, the elements of a will always be initialized (never have indeterminant value)? ie Even if T is a primitive type like int.
c++ c++17
asked 5 hours ago
Andrew Tomazos
34.6k25132227
Suppose I have a class template like this:
template<typename T, size_t N>
struct S {
std::array<T,N> a;
};
Is there a default member initializer I can place on a:
template<typename T, size_t N>
struct S {
std::array<T,N> a = ???;
};
such that no matter what T is, the elements of a will always be initialized (never have indeterminant value)? ie Even if T is a primitive type like int.
c++ c++17
c++ c++17
asked 5 hours ago
Andrew Tomazos
34.6k25132227
asked 5 hours ago
Andrew Tomazos
34.6k25132227
asked 5 hours ago
Andrew Tomazos
34.6k25132227
asked 5 hours ago
Andrew Tomazos
34.6k25132227
asked 5 hours ago
Andrew Tomazos
34.6k25132227
34.6k25132227
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
This:
template<typename T, size_t N>
struct S {
std::array<T,N> a = {};
};
That will recursively copy-initialize each element from {}. For int, that will zero-initialize. Of course, someone can always write:
struct A {
A() {}
int i;
};
which would prevent i from being initialized. But that's on them.
answered 5 hours ago
Barry
177k18304560
add a comment |
std::array is an aggregate type. You can aggregate initialize it with empty braces {} and that will initialize accordingly the elements of the internal array of T that std::array holds.
answered 5 hours ago
Jans
8,46522635
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
This:
template<typename T, size_t N>
struct S {
std::array<T,N> a = {};
};
That will recursively copy-initialize each element from {}. For int, that will zero-initialize. Of course, someone can always write:
struct A {
A() {}
int i;
};
which would prevent i from being initialized. But that's on them.
answered 5 hours ago
Barry
177k18304560
add a comment |
This:
template<typename T, size_t N>
struct S {
std::array<T,N> a = {};
};
That will recursively copy-initialize each element from {}. For int, that will zero-initialize. Of course, someone can always write:
struct A {
A() {}
int i;
};
which would prevent i from being initialized. But that's on them.
answered 5 hours ago
Barry
177k18304560
add a comment |
This:
template<typename T, size_t N>
struct S {
std::array<T,N> a = {};
};
That will recursively copy-initialize each element from {}. For int, that will zero-initialize. Of course, someone can always write:
struct A {
A() {}
int i;
};
which would prevent i from being initialized. But that's on them.
answered 5 hours ago
Barry
177k18304560
This:
template<typename T, size_t N>
struct S {
std::array<T,N> a = {};
};
That will recursively copy-initialize each element from {}. For int, that will zero-initialize. Of course, someone can always write:
struct A {
A() {}
int i;
};
which would prevent i from being initialized. But that's on them.
answered 5 hours ago
Barry
177k18304560
edited 4 hours ago
answered 5 hours ago
Barry
177k18304560
answered 5 hours ago
Barry
177k18304560
answered 5 hours ago
Barry
177k18304560
177k18304560
add a comment |
add a comment |
std::array is an aggregate type. You can aggregate initialize it with empty braces {} and that will initialize accordingly the elements of the internal array of T that std::array holds.
answered 5 hours ago
Jans
8,46522635
add a comment |
std::array is an aggregate type. You can aggregate initialize it with empty braces {} and that will initialize accordingly the elements of the internal array of T that std::array holds.
answered 5 hours ago
Jans
8,46522635
add a comment |
std::array is an aggregate type. You can aggregate initialize it with empty braces {} and that will initialize accordingly the elements of the internal array of T that std::array holds.
answered 5 hours ago
Jans
8,46522635
std::array is an aggregate type. You can aggregate initialize it with empty braces {} and that will initialize accordingly the elements of the internal array of T that std::array holds.
answered 5 hours ago
Jans
8,46522635
edited 4 hours ago
answered 5 hours ago
Jans
8,46522635
answered 5 hours ago
Jans
8,46522635
answered 5 hours ago
Jans
8,46522635
8,46522635
add a comment |
add a comment |
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