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assembly - 如何将参数传递给ARM汇编语言？

问题描述

我有以下 A76 内核的汇编语言程序。如您所见，我加载到 X0 的值是硬编码的。如何传递参数并在此程序的循环中运行代码：

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

X0 (0x000000001800009B) 的值是寄存器的编码地址，其中包含缓存方式和索引等变量。我必须更改循环中的方式和索引以从 C 代码中读取此寄存器的不同内容，如下所示：

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

另一个问题是：如果我像在汇编代码中那样读取 X0 并从函数返回，那么该值会转到 I0_W0 吗？

标签： assemblyarmdisassembly

解决方案

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.

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