I encountered a variant of this code when looking at another question (the original code used a std::thread
instead of std::vector
, but the syntax is the same):
#include <iostream>
#include <vector>
#include <iterator>
#include <algorithm>
int main()
{
std::vector<double> vecs[10] = std::vector<double>(10, 1);
for(auto& vec: vecs){
std::copy(vec.begin(), vec.end(), std::ostream_iterator<double>(std::cout, " "));
std::cout<<std::endl;
}
return 0;
}
This code shouldn't compile; std::vector<double> vecs[10] = std::vector<double>(10, 1);
is not valid initialization syntax, and clang rejects it with error: array initializer must be an initializer list
. However, GCC accepts it and appears to initialize every vector in the list with a copy of the specified temporary.
Is this some GCC extension I've never heard about (that somehow also managed to survive -pedantic-errors
) or just a plain bug?
I would consider this a bug.
#include <vector>
int main()
{
std::vector<double> x = std::vector<double>(10, 1);
std::vector<double> vecs[10] = x;
return 0;
}
Works (as you have spotted).
While
int main()
{
int x = 10;
int is[10] = x;
return 0;
}
yields the (expected) error.
Further investigation:
struct A { A() { } };
int main() { A a[10] = A(); }
This compiles in GCC.
struct A { A() = default; };
int main() { A a[10] = A(); }
Also compiles in GCC 4.9 but not earlier versions I tested (4.6-4.8).
struct A { };
int main() { A a[10] = A(); }
Doesn't compile.
struct B { virtual ~B() { } };
struct A : B { };
int main() { A a[10] = A(); }
Compiles.
struct B { ~B() { } };
struct A : B { };
int main() { A a[10] = A(); }
Doesn't compile.
I think it's safe to say that it's a bug. No sane extension would have this sort of behavior. Note that A
in both the second case and the third case are POD types (the only difference being an explicitly defaulted default constructor), but they are treated differently by GCC 4.9.
Edit: Going through the GCC bugzilla again, this bug report appears to be related. Looks like I missed it the first time through because the title was talking about string literals.
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