From: Lars Schouw (schouwla_at_[hidden])
Date: 2008-05-21 03:16:19
Thanks Joaquin I will try boost::array.
----- Original Message ----
From: "joaquin_at_[hidden]" <joaquin_at_[hidden]>
To: "boost_at_[hidden]" <boost_at_[hidden]>
Sent: Wednesday, May 21, 2008 4:05:50 PM
Subject: Re: [boost] boost::get<i>
Lars Schouw escribió:
> I tried to use boost::tuple for the first time, and soon ran into a problem.
> I can't get values using boost::get<i> where i is an int.
> samplec code
> typedef boost::tuple<double,double,double> tuple;
> void foo()
> std::vector<tuple> corr = boost::assign::tuple_list_of ( 1.0, 0.1, -0.1 ) ( 0.1, 1.0, 0.4 ) ( -0.1, 0.4, 1.0 );
> for (int i = 0; i < 3; i++)
> for (int j = 0; j < 3; j++)
> double val = boost::get<i>( corr[j] );
> Using Visual Studio 2005 this comes give me a compiler error:
> error C2971: 'boost::tuples::get' : template parameter 'N' : 'i' : a local
> variable cannot be used as a non-type argument
> then when I move the int i declaration outside of the function foo.
> I get a new error:
> error C2975: 'N' : invalid template argument for 'boost::tuples::get',
> expected compile-time constant expression
> What do I need to do to access values in a tuple with an index i?
The problem is that you can't do that for very fundamental reasons. As
complains about, i in get<i> must be a compile-time constant, not some
value that can change
during the execution of the program. To see why, consider the case where
type had been defined as:
// component 1 is not a double bout a char*
typedef boost::tuple<double,char*,double> tuple;
Now, take a look again at the expression:
double val = boost::get<i>(corr[j]));
What happens when i==1? The compiler cannot know in advance what values i
will take, so it demands that i be a compile-time constant so as to be
able to do its
In your particular case where all the component types are the same, you
replacing the tuple with boost::array.
Joaquín M López Muñoz
Telefónica, Investigación y Desarrollo
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