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support multiplication of two float3 operands

Open sahooora opened this issue 4 years ago • 2 comments

Hi,

Thanks for your work.

I have a C code with variables in float3 data type. So I included float3.h header file in my code and compiled it with g++. It seems that the library doesn't support the multiplication of two float3 operands, am I right?

For example for the following line: ((float3*)((float*)compute1 + (0)))[0] = (((float3*)((float*)compute1 + (0)))[0] + (((float3*)((float*)placeholder + (0)))[0] * ((float3*)((float*)placeholder1 + ((y_outer_x_outer_fused * 3))))[0]));

I get this error:

error: no match for ‘operator*’ (operand types are ‘math::float3’ and ‘math::float3’)
       ((float3*)((float*)compute1 + (0)))[0] = (((float3*)((float*)compute1 + (0)))[0] + (((float3*)((float*)placeholder + (0)))[0] * ((float3*)((float*)placeholder1 + ((y_outer_x_outer_fused * 3))))[0]));
                                                                                           ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

sahooora avatar Jan 07 '21 13:01 sahooora

That is true, float3 * float3 is not supported by default. The rationale behind that back in the time was that point-wise multiplication is not a mathematically defined operation in linear spaces, and would possibly be a programming bug. There is a #define MATH_ENABLE_UNCOMMON_OPERATIONS that you can enable at command line to enable float3 * float3.

See https://github.com/juj/MathGeoLib/blob/2940b99b99cfe575dd45103ef20f4019dee15b54/src/Math/float3.h#L155

Instead, by default the more explicit form oneFloat3.Mul(anotherFloat3) is used, so to be explicit when point-wise multiplication is taking place.

In hindsight it is possible that no confusion would arise from enabling it by default, but back then that was the thinking.

juj avatar Jan 07 '21 19:01 juj

Thanks. It's solved.

sahooora avatar Jan 07 '21 20:01 sahooora