10 years, 4 months ago.

Float precision and implicit conversion, why?

Hello all,

I'm seeing some warnings that i don't fully understand.

I have Pi and a constant float defined as follows:

#ifndef M_PI
#define M_PI           3.1415927
#endif

const float COXA_ANGLE  = M_PI/4.0; //45deg;     // Angle of coxa from straight forward (deg)

If i use:

leg[1].footPosCalc.x = tempFootPosY[1]*cos(COXA_ANGLE*2.0) - tempFootPosX[1]*sin(COXA_ANGLE*2.0);

I get warnings like: Warning: Single-precision operand implicitly converted to double-precision in "ik.cpp", Line: 147, Col: 41

If i use:

leg[1].footPosCalc.x = tempFootPosY[1]*cos(COXA_ANGLE*2) - tempFootPosX[1]*sin(COXA_ANGLE*2);

I get no warnings.

I understand that in the first case i'm using 2.0 which is a float, but since its being multiplied against a float, isn't the result another float?

In the multiply by integer '2' case, isn't the two being cast into a float for the multiplication anyway?

In either case, why is it converting to a double precision float? I'm happy with the answer being single precision.

I'm using an LPC4088 and my understanding is that the FPU only works for 32bit single precision floats so i want to make sure i stay in that domain. Is this true and if so am I on the right track?

Thanks a lot, Dan