IQmath howto divide long Values

Hello!

If you use C28x refer to

 http://processors.wiki.ti.com/index.php/IQmath_Library_for_C28x#Q:_If_I_use_IQmath_f32.lib.2C_does_it_do_native_floating_point_operations.3F

In general, for the C-language your problem is solved thus:

solution = float(A) / float(B);

Regards,

Igor

Hi Stefan

If your MCU have a FPU the proposal of Igor is nice. If not maybe this way can help you.

First be sure in which range your variable A and B can be. Table with range of IQMATH types you can see in SPRC990, p. 10. In your example the variables would fit in a IQ11 type.

A =<<11; // transform type to IQ11

B=<<11; // transform type to IQ11

solution = _IQdiv11(A, B);

Regards

Matt

Hi!

Agree, "solution = float(A) / float(B);" will not work without FPU.

Regards,

Igor 

Hi!

In your operations "helpvar=_IQ11div(A,B);" involving only integer types. I think you're missing the mantissa. It seems your result is "solution= XXXX.0".

Regards,

Igor

No, i am getting the nearly correct value of 0.2705078 for my example.  

Igor Gorbachev said:

Hi!

In your operations "helpvar=_IQ11div(A,B);" involving only integer types. I think you're missing the mantissa. It seems your result is "solution= XXXX.0".

Regards,

Igor

An IQ value is stored in a long.  _iq is simply a typedef of long in the IQ header file:

//===========================================================================
typedef   long    _iq;
typedef   long    _iq30;
typedef   long    _iq29;
typedef   long    _iq28;
typedef   long    _iq27;
typedef   long    _iq26;
typedef   long    _iq25;
typedef   long    _iq24;
typedef   long    _iq23;
typedef   long    _iq22;