FPU: CubeIDE setting overrides register SCB->CPACR

[LCE]

Heyho,

just played with / checked the FPU on a G431.

I found that the CubeIDE setting overrides the enable via register SCB->CPACR.

So, if source code sets FPU on, but IDE says FPU off, the FPU is off.

That a bug or a feature?

[mƎALLEm]

So if it’s a question on CubeIDE, that must be asked in STM32CubeIDE.

I’m moving the post then.

[TDK]

That’s not how it works. The IDE can set some things up but once the source code is running it controls things. Recheck your assumptions.

[LCE]

Okay, let’s rephrase it:

something changes it, probably the compiler ?

[waclawek.jan]

> if source code sets FPU on

How?

> but IDE says FPU off, the FPU is off

How do you know FPU is off?

 

I don’t use CubeIDE, but my guess would be, that any “FPU on/off” tickbox results in preprocessor symbols being defined/not defined, probably through command-line -Dxxx switch, which are then used in SystemInit()

void SystemInit (void)
{
  /* FPU settings ------------------------------------------------------------*/
  #if (__FPU_PRESENT == 1) && (__FPU_USED == 1)
    SCB->CPACR |= ((3UL << (10*2))|(3UL << (11*2)));  /* set CP10 and CP11 Full Access */
  #endif

so if your code changes that before SystemInit is called, it will be overriden.

Or not. I don’t Cube.

JW

[LCE]

Ouch, feeling stooopid once more…

I copied exactly the above lines including the preprocessor #if, without having checked where the __FPU defines come from. I assumed wrongly that these were set in some *.h file.

Ouch, ouch, ouch, ...

[LCE]

BTW, I checked by setting up a timer running with CPU clock and checked the timer ticks.

That showed pretty clearly if FPU was on or not, a division taking about 10 times more cycles without the FPU.

case '/':
case ':':
{
	__disable_irq();
	TIM6->CNT = 0;
	if( flValY != 0.0f ) flResult = flValX / flValY;
	else u32Error = 1;
	__ISB();
	u32TimStop = TimTestGetCnt();
	__enable_irq();
	break;
}