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标题: TL138 uboot是怎么区分配置两款核心板的 [打印本页]

作者: Mr.Loser 时间: 2014-9-11 09:33
标题: TL138 uboot是怎么区分配置两款核心板的
核心板1：DDR2 128M Byte NAND FLASH 4G bit
核心板2：DDR2 256M Byte NAND FLASH 8G bit
这两个核心板公共一个uboot，DDR2的配置肯定是不同的，uboot怎么检查到是哪个核心板，然后去执行相应的配置呢？

是不是通过读NAND FLASH的ID，两款核心板NAND FLASH不同，ID也不同，这样读到了ID就知道是哪个核心板了？

作者: 2532609929 时间: 2014-9-11 18:55
DDR通过检查最大有效地址来识别容量，具体请看uboot的common/memsize.c文件中的检查代码，也可以看如下：
/*
* Check memory range for valid RAM. A simple memory test determines
* the actually available RAM size between addresses `base' and
* `base + maxsize'.
*/
long get_ram_size(long *base, long maxsize)
{
      volatile long *addr;
      long          save[32];
      long          cnt;
      long          val;
      long          size;
      int          i = 0;

      for (cnt = (maxsize / sizeof (long)) >> 1; cnt > 0; cnt >>= 1) {
            addr = base + cnt;       /* pointer arith! */
            sync ();
            save[i++] = *addr;
            sync ();
            *addr = ~cnt;
      }

      addr = base;
      sync ();
      save = *addr;1 h( f3 P& _6 T2 C. R! L
      sync ();- u9 k2 l& W7 T  c
      *addr = 0;
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' N+ f8 M5 C( w6 o4 U       sync ();# R5 J. z& O* A6 D
      if ((val = *addr) != 0) {
- x- C8 i2 f* s6 h             /* Restore the original data before leaving the function.5 u1 G6 q* R' E# ^# ^
               */
, M* j4 \9 @% Z) b: ~             sync ();6 z7 k9 U0 Z! a" V4 x
            *addr = save;
) S+ X, e' O' w+ @+ Y             for (cnt = 1; cnt < maxsize / sizeof(long); cnt <<= 1) {$ X6 ?2 ]4 I& K- q) B8 k" C
                     addr  = base + cnt;$ Q/ T7 ^3 K, {" l
                     sync ();% d, E/ ~. z" n7 h, V
                     *addr = save[--i];
) ~+ @- z% D  U) o8 G* u             }' f" j3 X6 R% z/ B. ~" x# w; @% G
            return (0);4 f* R, m; k$ t: J
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      for (cnt = 1; cnt < maxsize / sizeof (long); cnt <<= 1) {
# ?  y; Q0 M' A9 w* z             addr = base + cnt;       /* pointer arith! */
  Q0 Q1 D. O$ m% \3 J! V8 N             val = *addr;& e6 N+ c% H- W" j
            *addr = save[--i];
0 L4 q! ?# j0 ~% f7 k             if (val != ~cnt) {
$ J# q% T  P1 N% O                      size = cnt * sizeof (long);* D5 l8 G' h) t3 q) _
                     /* Restore the original data before leaving the function.7 V4 p1 V/ x! y, B8 @- \
                     */$ p- B4 b7 e, I2 T7 [3 v
                     for (cnt <<= 1; cnt < maxsize / sizeof (long); cnt <<= 1) {2 n! L$ G# M" N- ~+ S7 a7 m) R
                              addr  = base + cnt;
# y( Z# F/ y0 E! d( ^                               *addr = save[--i];
7 a9 v) U: r+ X6 l; n- P                      }
( B) s$ P. t8 ?& W5 |7 y! F8 H                      return (size);  Z7 P# Q: B/ r
            }1 g2 n* K0 x" g1 z2 ?% s
      }
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      return (maxsize);4 s/ u# b& l, _2 M! ]% S9 n7 V4 z- G
}6 w# X5 h6 z' F+ C* D& ?  v% q
int dram_init(void)8 v# F' s, y* k6 W- ~
{
1 T6 L: X9 ~0 d& X+ A% V       /* dram_init must store complete ramsize in gd->ram_size */
1 G! ^/ x" ?3 u       gd->ram_size = get_ram_size(  Z. z' t7 x% f) l) P) B
                     (void *)CONFIG_SYS_SDRAM_BASE,
; ?5 S% @8 b& `: D  B1 t0 P# ?                      CONFIG_MAX_RAM_BANK_SIZE);
& m9 _3 {8 v. P2 R       return 0;
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+ I, q7 c6 t" o5 K2 J5 U; [FLASH是通过检查FLASH内部的ID识别容量，希望对您有帮助！% t9 P# `  s2 {" j/ G  l2 {, }- q3 t
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