Re: size of array is not an integral constant-expression
johnehein@gmail.com wrote:
#include <vector>
using namespace std;
template <typename Iter>
int
foo(Iter first, Iter last, int nn)
{
const size_t n = last - first;
double buf[n];
return 0;
}
int
main(int argc, char **argv)
{
vector<double> x;
foo(x.begin(), x.end(), argc);
return 0;
}
foo.cc:17: instantiated from here
foo.cc:9: error: size of array is not an integral constant-expression
g++ 4.2.1
Is this error specific to g++ 4.x?
Not that I can see. 'last - first' is a run-time expression. And when
you use it to initialise a 'const size_t', the variable ('n') also
becomes a run-time expression. It cannot be used to declare an array.
g++ 3.6.4 and g++ 2.9.5 have no
problems with it, but that doesn't mean they are right. Is there some
reason to expect this to fail.
Yes, there is. It goes against the rules of the language. The older
versions of G++ may have had it as an extension. Hell, the new versions
may still have it as an extension, and you're welcome to use it, just
don't claim your program to be C++.
There are a few interesting workarounds that point to this being
unexpected behavior... I'll post those next.
What's so unexpected in actually implementing the rules of the language?
V
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