[boost] ublas int*vector != double*vector?

[Julius Muschaweck]

Hi,

this is my first posting, so please give me guidance if this is going into the wrong direction.

I'm using ublas (with Visual Studio .NET) , and I like it a lot.
However, in scalar * vector operations, ublas doesn't handle implicit int - double conversions like the compiler, which resulted in a hard to trace bug for me:

using namespace boost::numeric::ublas;
vector<double> v(3);
v[0] = 1.1;
v[1] = 2.2;
v[2] = 3.7;
vector<double> v2;
v2 = 2.0 * v; // double times vector
std::cout << v2 << std::endl;
v2 = 2 * v; // integer times vector
std::cout << v2 << std::endl; //oops
std::cout << 2*3.7 << ' ' << 2.0 * 3.7 << std::endl;

produces

(2.2,4.4,7.4)
(2.,4.,7.)
7.4 7.4

What seems to happen is that ublas first rounds the doubles in v to ints, performs integer multiplication and then converts the result back to double, whereas the compiler converts an int to double and then performs floating point multiplication.

It's maybe not good style to rely on implicit conversions like this, but I'm surely not the only one.

My opinion is that ublas should do int*vector<double> like the compiler does int*double, or am I missing a reason why this is intended?

Julius Muschaweck

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