CODEWITHSUNDEEP
assemblyarmdisassembly
user3565150
user3565150

Reputation: 914

How to pass parameters to ARM assembly language?

I have the following assembly language program for A76 core. As you can see that the value I am loading to X0 is hard coded. How can I pass parameters and run the code in a loop for this program:

EXPORT tlb_ram_data_register_18_1
tlb_ram_data_register_18 PROC
;   B asm_Switch2NonSecure  
;   B EL1_to_EL3_Switch
LDR X0, =0x000000001800009B
SYS #6, c15, c0, #0, X0
DSB SY
ISB
MRS X0, S3_6_c15_c0_1 ; Move ILData0 register to X1
RET 
ENDP

The value for X0 (0x000000001800009B) is the encoding address of the register which contains variables like cache way and indices. I have to change the way and indices in loop to read different contents of this register from a C code like below:

for(way0 = 0x0;way0 <= 0xFF ; way0++)
{
    I0_W0 = tlb_ram_data_register_18_1(way0);
}

One more question is: If I read X0 like I am doing in the assembly code and return from the function, then will that value go to I0_W0?

Upvotes: 1

Views: 4029

Answers (1)

old_timer
old_timer

Reputation: 71536

technically it is compiler specific, but for ARM a number of them conform to the same ABI. But you can just ask the compiler and it will give you the correct answer. Write a program test program

extern unsigned int tlb_ram_data_register_18_1 ( unsigned int );
unsigned int fun ( void )
{
    unsigned int way0;

    for(way0 = 0x0;way0 <= 0xFF ; way0++)
    {
        tlb_ram_data_register_18_1(way0);
    }
    return(5);
}

for the moment I am building an aarch64 compiler but for now here is what arm looks like, you will see something similar with any target.

I am using a current version of gcc

00000000 <fun>:
   0:   e92d4010    push    {r4, lr}
   4:   e3a04000    mov r4, #0
   8:   e1a00004    mov r0, r4
   c:   e2844001    add r4, r4, #1
  10:   ebfffffe    bl  0 <tlb_ram_data_register_18_1>
  14:   e3540c01    cmp r4, #256    ; 0x100
  18:   1afffffa    bne 8 <fun+0x8>
  1c:   e3a00005    mov r0, #5
  20:   e8bd4010    pop {r4, lr}
  24:   e12fff1e    bx  lr

not so obvious but r0 is being used as the parameter to the function

also note r0 is used for the return value

unsigned int fun ( unsigned int x )
{
    return(x+1);
}


00000000 <fun>:
   0:   e2800001    add r0, r0, #1
   4:   e12fff1e    bx  lr

r0 is both the first/only parameter and is also used for the return.

ahh I do have an aarch64 gcc handy

0000000000000000 <fun>:
   0:   11000400    add w0, w0, #0x1
   4:   d65f03c0    ret

w0 on the way and on the way out. you can try this again with a 64bit parameter and see if that changes.

Upvotes: 2

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