Is the second parameter in ioremap() gives the size in number of bits for a register- Linux?

Question

My NEC microcontroller has a timer controller register 8-bits - Do, I need to pass 8 in the second parameter of ioremap?

After reading the spec, I got to know the following property of it.

Address    |Function Register Name   |Symbol   |R/W Manipulatable Bits   |Default Val

FFFFF590H  |TMP0 control register 0  |TP0CTL0  |R/W √ √                  |00H

So, I believe that the physical address at which the Timer register TP0CTL0 is mapped is 0xFFFFF590.

Now, I am remapping this register as the following. After reading more description, I got to know that the register is 8-bit in size.

The spec says "The TPnCTL0 register is an 8-bit register that controls the operation of TMPn."

Is this right? I am using the base address as 0xFFFFF590 and the size of this register is 8-bits. Thus, I have given the size as 8-bit. Is it correct? Is the second paramter of ioremap_nocache is in the size of bits? Is my following API is correct? Have I used the parameters correctly in the function - ioremap_nocache.

void *tp0ctl0 = ioremap_nocache(0xFFFFF590, 8);

Next, I am doing the following -

unsigned int val =  ioread8(tp0ctl0);
val = 2;
iowrite8(val, tp0ctl0);

Please correct me here. Please let me know, if I am using the API's correctly or not based on the microcontroller information I have.

Answer 1

The size given to ioremap_* is in bytes not bits. The purpose of this function is to map physical address space into kernel virtual address though, so anything greater than zero and less than or equal to the system page size will be equivalent.

Given the information you provided above, ioremap_nocache(0xFFFFF590, 1) would actually be correct. But the effect of "1" versus "8" will be identical since the system page size is (no doubt) larger than both.