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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;
$ C1 y) V$ j0 K" f8 B- O& _' @       sync ();
; Y& a& K  H/ s) d$ H2 z' [1 P       *addr = 0;
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5 H+ e$ G" J( u       sync ();& F" j  L# b% e) E0 c7 {" L
      if ((val = *addr) != 0) {
$ C. |( Q% `! A9 c1 X/ Q. I             /* Restore the original data before leaving the function.
. D* {0 Q* f, {6 ?+ R: E$ X                */
1 c$ R% y* S8 Z- t3 y9 F+ M             sync ();
" V+ q3 _6 D0 B3 g- L' V& L' I& p& t             *addr = save;
9 q. \, g; T3 B# L% h             for (cnt = 1; cnt < maxsize / sizeof(long); cnt <<= 1) {. }- [8 [) P. N. P
                     addr  = base + cnt;- S: a, u# b, o- U+ d" S
                     sync ();
/ `3 M) `. d' n2 I, g. N0 [6 N                      *addr = save[--i];: G% Y0 d/ m, d4 L
            }8 _! q) P1 \& y# `, z
            return (0);6 ?4 i$ G2 r" F( Z/ g
      }% [# ~9 h( z, h
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      for (cnt = 1; cnt < maxsize / sizeof (long); cnt <<= 1) {
$ a3 V1 K& ]' K9 R             addr = base + cnt;       /* pointer arith! */# @! P5 w$ n1 r" c1 X
            val = *addr;( N- n+ \6 p) ~  K9 k
            *addr = save[--i];8 x0 L$ ]( O  h7 s1 S6 n
            if (val != ~cnt) {
6 ~- s5 @& `( `0 q# t+ H) d/ \                      size = cnt * sizeof (long);+ k+ I$ h2 [# q
                     /* Restore the original data before leaving the function.
* D) s# i* g0 Y                      */* ^/ C- }4 i4 v& D* J
                     for (cnt <<= 1; cnt < maxsize / sizeof (long); cnt <<= 1) {
1 z$ ~/ P6 H# a$ s7 j                               addr  = base + cnt;& `& @, d. L4 }: b0 C' v
                              *addr = save[--i];
" M& q5 q& X3 w# R. z9 X1 f$ J                      }/ [5 W9 a7 @3 {3 D: \
                     return (size);
2 }' N% u9 F! n; _             }2 ~# f4 W9 C* _8 c
      }
. {+ N: R" H& R& ]) h* |9 r2 U1 |  S; s! ^
      return (maxsize);2 C! f' D: x9 E5 K, e
}
$ R, j- d" V1 u1 aint dram_init(void)# w. U$ l, C9 D# l- J/ B) I% j
{
- s" A+ }. w' \2 i* x& Y) F3 d       /* dram_init must store complete ramsize in gd->ram_size */  T6 r- C6 `+ U
      gd->ram_size = get_ram_size(+ P& u& x! O/ z1 i; o- g
                     (void *)CONFIG_SYS_SDRAM_BASE,' O4 j1 t( m, q# {
                     CONFIG_MAX_RAM_BANK_SIZE);
: }8 v+ T/ K# G7 x  b, ~       return 0;, q3 N6 ~! I* L7 `0 Q7 Q* K
}
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- U3 Y9 ]3 L2 x# [- q, G, xFLASH是通过检查FLASH内部的ID识别容量，希望对您有帮助！
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