GPIO speed depending on how I call printf before the loop?

HiFive1 Rev B
1

I have this very small assembly program that toggles a GPIO pin. Before the tight loop that toggles the GPIO pin I have a printf statement. Depending on how I call this printf statement the frequency of the GPIO toggling changes.

When I call print with call printf I get a frequency of 615 kHz and when I call printf with jal ra, printf I get a frequency of 571 kHz.

Can someone please, explain this to me?

This is my code (called from main.c):

.equ	GPIO,		0x10012000
.equ	OUTPUT_EN,	0x08
.equ	OUTPUT_VAL,	0x0c
.equ	PIN,		22		# the red led


assembler:
	la		a0, hello
	jal		ra, printf

	# setup register to blink
	la		t0, GPIO
	li		t1, 1
	li		t3, PIN

	# enable the specified GPIO pin
	lw		t4, OUTPUT_EN(t0) 		# get the current value of the enable bits
	sll		t2, t1, t3
	or	 	t4, t4, t2
	sw		t4, OUTPUT_EN(t0)

	sll 	t2, t1, t3
100:
	# toggle the GPIO pin
	lw 		t4, OUTPUT_VAL(t0)
	xor 	t4, t4, t2
	sw		t4, OUTPUT_VAL(t0)
	j		100b

	ret

		.section	.rodata

	hello:
		.string	"This is the clockspeed demo\r\n"
2

Has nobody an answer to this question?

Regards,
Bengt

3

It could be an alignment issue (of the jump target).

4

Thank you for your kind answer.

Could you, please, explain how that would affect the loop as the call to printf is done before the loop? I don’t understand this.