John,
In the meantime, consider the following workaround:
std::ostringstream o;
o << std::fixed << std::setprecision(1) << tmp;
int128_t roundedNanoseconds(o.str().substr(0,o.tellp() - std::streampos(2)));
Does that seem reasonable, or do you see a much cleaner or efficient approach?
Would be nicer to use setprecision(0), but:
https://svn.boost.org/trac/boost/ticket/10103
Thanks for your input,
Rob Conde
________________________________
From: Rob Conde
Sent: Saturday, September 17, 2016 3:57:39 PM
To: boost-users@lists.boost.org
Subject: Re: [Boost-users] [boost][multiprecision] compile error with gcc 4.8.4
Understood...thanks for your effort.
Rob Conde
Sent from my iPhone
On Sep 17, 2016, at 2:43 PM, John Maddock mailto:jz.maddock@googlemail.com> wrote:
On 17/09/2016 18:35, Rob Conde wrote:
The following works with me in visual studio with toolset v120 and v140.
#include
#include
#include <iostream>
typedef boost::multiprecision::int128_t int128_t;
typedef boost::multiprecision::cpp_bin_float_quad float128_t;
void main()
{
float128_t floatVal("6.0");
int128_t iVal(floatVal.convert_to());
std::cout << iVal << std::endl;
}
If I change it to have a fractional component, then indeed I get an exception...maybe that's what you mean. I personally only need the above behavior (though keeping the c++ truncation semantics might make sense as a long term goal...but i digress). I always floor/ceil because call convert_to().
Using boost 1.61 and gcc version 4.4.7, the above also compiles and works.
OK, I see that compiles with 1.61, it does not with current master though (what will be 1.62).
In 1.61 the code performs a lexical_cast for the conversion - that works more or less by accident for small integers, but for large ones it fails at runtime (as an exception) as the numbers get printed in scientific format during the lexical_cast. Conversion to long long works fine though, and probably in all the cases that your conversion to int128_t does.
As I said, I need to figure out some more consistent behaviour for future releases.
John.
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