汇编对伪代码的理解

Question

I am trying to reverse a executable file (for learning RE), However i am stuck on a piece of pseudocode which has some things that I dont understand.Kindly help me to understand it so I can improve. Now I understand that there is an array which has number 32 till 2014th entry and from 2014th to 4095th entry there are the codes of character returned by fget.After that there is a little snippet of code which I am not able to understand,the block is marked with **.

Code:

int sub_12A11A0()
{
  __int16 v1; // [sp+4h] [bp-24h]@13
  int v2; // [sp+8h] [bp-20h]@11
  int v3; // [sp+Ch] [bp-1Ch]@8
  signed int v4; // [sp+10h] [bp-18h]@13
  signed int l; // [sp+14h] [bp-14h]@16
  int v6; // [sp+18h] [bp-10h]@6
  int v7; // [sp+18h] [bp-10h]@13
  int v8; // [sp+18h] [bp-10h]@32
  signed int v9; // [sp+1Ch] [bp-Ch]@8
  signed int v10; // [sp+20h] [bp-8h]@8
  signed int i; // [sp+24h] [bp-4h]@1
  signed int j; // [sp+24h] [bp-4h]@4
  int k; // [sp+24h] [bp-4h]@13
  signed int m; // [sp+24h] [bp-4h]@27

  for ( i = 0; i < 2015; ++i )
    byte_12A3400[i] = 32;     
  for ( j = 2015; j < 4096; ++j )
  {
    v6 = fgetc(dword_12A33EC);
    if ( v6 == -1 )
      break;
    byte_12A3400[j] = v6;
    ++dword_12A3088;
  }
  v9 = j;
  v10 = 2015;
  v3 = 0;
  while ( v10 < v9 )
  {
    if ( v9 - v10 < 33 )
      v2 = v9 - v10;
    else
      v2 = 33;
    v1 = 0;
    v4 = 1;
    ****v7 = (unsigned __int8)byte_12A3400[v10];
    for ( k = v10 - 1; k >= v3; --k )
    {
      if ( (unsigned __int8)byte_12A3400[k] == v7 )
      {
        for ( l = 1; l < v2 && (unsigned __int8)*(&byte_12A3400[l] + k) == (unsigned __int8)*(&byte_12A3400[l] + v10); ++l )
          ;
        if ( l > v4 )
        {
          v1 = k;
          v4 = l;
        }
      }
    }****
    if ( v4 > 1 )
      sub_12A1120(v1 & 0x7FF, v4 - 2);
    else
      sub_12A10D0(v7);
    v10 += v4;
    v3 += v4;
    if ( v10 >= 4063 )
    {
      for ( m = 0; m < 2048; ++m )
        byte_12A3400[m] = byte_12A3C00[m];
      v9 -= 2048;
      v10 -= 2048;
      v3 -= 2048;
      while ( v9 < 4096 )
      {
        v8 = fgetc(dword_12A33EC);
        if ( v8 == -1 )
          break;
        byte_12A3400[v9++] = v8;
        ++dword_12A3088;
      }
    }
  }
  return sub_12A1090();
}

Answer 1

It seems the important part is to understand

*(&byte_12A3400[l] + k)

This is equivalent to

byte_12A3400[l + k]