AGAL Opcode mul and dp4

Question

I am trying to dive into code AGAL and have some questions which are interesting for me.

The first question is about opcodes mul and dp4
If I am right, mul does the following

mul ft0, ft0, ft5 means ft0 = ft0 * ft5 which in fact is ft0.r*ft5.r +ft0.g*ft5.g+ft0.b*ft5.b+ft0.a*ft5.a

dp4 ft0, ft0, ft5 does the same as mul, but you can change an order by using another component order, for instance ft0.argb?
Thank you in advance!

Answer 1

No, you wrong. mil means ft0 = ft0 * ft5 = [ft0.r*ft5.r, ft0.g*ft5.g, ft0.b*ft5.b, ft0.a*ft5.a]. You get a vector. dp4 returns scalar (ft0.r*ft5.r + ft0.g*ft5.g + ft0.b*ft5.b + ft0.a*ft5.a). And yes, you can change an "order" - it's called swizzling.