Using gcc-4.1.1. Info says variable-length array is supported in c++ mode,
but doesn't seem to work:
#include
#include
template
void F (in_t const& in, int size, int x[size]) {}
void G (std::vector const& in, int size, int x[size]) {}
int main () {
std::vector i (10);
int x (10);
F (i,
gcc-4.3.0-8.x86_64
I have test code that does passes std::isfinite (x), yet if I print the
values to std::cout the value printed is 'inf'. Is std::isfinite (x)
broken?
Paolo Carlini wrote:
> Neal Becker wrote:
>> gcc-4.3.0-8.x86_64
>>
>> I have test code that does passes std::isfinite (x), yet if I print the
>> values to std::cout the value printed is 'inf'. Is std::isfinite (x)
>> broken?
>>
> Whatever
Paolo Carlini wrote:
> ... ah, ok, now I see what you meant, you meant that x is *not* finite,
> still, std::isfinite(x) != 0. Still, testcase badly needed...
>
> Paolo.
#include
#include
int main () {
double x = log (0);
if (not std::isfinite (x)) {
throw std::runtime_error ("not fini
Paolo Carlini wrote:
> Hi ho, ho!! ;)
>> It worked with me.
>>
> Try a recent gcc (eg, 4.3.x) and you will get the same, actually
> expected, result of the original poster.
>
> Paolo.
I believe this is a bug. I agree that -ffast-math will not always comply 100%
with IEEE, as advertised. Bu
gcc version 4.1.2 20070502 (Red Hat 4.1.2-12)
I noticed the following code
=== version 1:
template
inline a_t append (a_t & a, b_t const& b) {
a.insert (a.end(), b.begin(), b.end());
return a;
}
=== version 2:
template
inline void append (a_t & a, b_t const& b) {
a.insert (a.end(), b.begin(
I need a preprocessor macro to detect c++0x support. For now, that is
__GXX_EXPERIMENTAL_CXX0X__
but what happens once -std=c++0x is the default? Will this macro still
be defined?
Don't we need a
__GXX_CXX0X__ ?
