arm - ARM v7 中未对齐的内存访问

Question

你能帮我用代码（'C'或ARM程序集）将内存区域标记为“正常”，从而允许未对齐的内存访问吗？我知道我们需要在执行此操作之前启用 MMU。我是 ARM 架构的新手。

谢谢！

Answer 1

如果您只需要非对齐访问，请尝试将 cp15 sctlr[1]（对齐位）设置为 0。

mrc  p15, 0, r0, c1, c0, 0
bic  r0, r0, #2
mcr  p15, 0, r0, c1, c0, 0 

我不知道是否需要为此启用 MMU。

Answer 2

#define L1PoniterTo2ndLevelPageTable(BASE_ADDRESS, P, DOMAIN) (BASE_ADDRESS<<10U | P<<9U | DOMAIN<<5U | 1U)
#define L1Section(BASE_ADDRESS, SBZ, NG, S, APX, TEX, AP, P, DOMAIN, XN, C, B) (BASE_ADDRESS<<20U | SBZ<<19 | 0U<<18U | NG<<17U | S<<16U | APX<<15U | TEX<<12U | AP<<10U | P<<9U | DOMAIN<<5U | XN<<4U | C<<3U | B<<2 |2U)
#define L1SuperSection(BASE_ADDRESS, SBZ, NG, S, APX, TEX, AP, P, DOMAIN, XN, C, B) (BASE_ADDRESS<<24U | SBZ<<19 | 1U<<18U | NG<<17U | S<<16U | APX<<15U | TEX<<12U | AP<<10U | P<<9U | DOMAIN<<5U | XN<<4U | C<<3U | B<<2 |2U)

unsigned long au32PageTableL1[4096U] __attribute__((aligned(0x4000)));
//extern unsigned long _endof_consts[4096U];
unsigned long u32EnableMMU()
{
  unsigned long u32Index_l;
  unsigned long dummy;
  unsigned long sctlr;

  //create L1 page table
  for (u32Index_l=0;u32Index_l<sizeof(au32PageTableL1)/sizeof(au32PageTableL1[0]);u32Index_l++)
  {
    au32PageTableL1[u32Index_l] = L1Section
        (
        u32Index_l, //BASE_ADDRESS
        0b0, //SBZ
        0b0, //NG
        0b0, //S
        0b0, //APX
        0b001, //TEX
        0b11, //AP
        0b0, //P
        0,//DOMAIN
        0b0, //XN
        0b1, //C
        0b1 //B
        );
  }

  //set ttbr0
  __asm("MCR p15, 0, %0,       c2, c0, 0" : : "r" (au32PageTableL1));

  //invalidate TLB
  __asm("MCR p15, 0, %0,       c8, c3, 0" : : "r" (dummy));

  /* Ensure all TLB maintenance operations complete before returning. */
  __asm("dsb");

  //enable MMU
  __asm("MRC p15, 0, %0,       c1, c0, 0" : "=r" (sctlr));
  sctlr |= 1U;
  __asm("MCR p15, 0, %0,       c1, c0, 0" : : "r" (sctlr));

  return 0;
}