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* _1 h* p$ g7 ?) e8 O/ d5 T
核心板2：DDR2 256M Byte NAND FLASH 8G bit ; s$ g8 g5 Q3 }0 ?- P V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？ , r1 Q! m% m; n, [% h; \ * b4 R1 e; y n9 p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？. a+ [# A' ~) N U$ r
8 k& Q K) Z) Q% A1 H) X% h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：5 t% C5 t4 N9 [
/* + V' U) M9 P4 U# b( b) V8 Y- P) T* Check memory range for valid RAM. A simple memory test determines . L& R# b$ O+ n- L" Z* the actually available RAM size between addresses `base' and 6 U$ ?& J: _ F1 y* `base + maxsize'. $ `/ q& @! l+ G, ^2 C/ O: }*/. W3 y* C9 c) }4 a
long get_ram_size(long *base, long maxsize)3 Q2 Q& q/ r0 ?; @0 H: Y2 s, h8 y
{% K6 C/ G" `: ^7 ~# |1 R& W
volatile long *addr; ^& l& E4 w2 u! C f7 a) f long save[32]; 2 V( G4 W. p M2 E8 j2 K2 D long cnt; : g" ?5 J" `) g5 \9 W- J) g, G7 s long val;0 t* l7 a. j, V4 n% g3 u
long size;. o8 U, d# i1 Y
int i = 0;* L6 S% [4 I3 n
0 y: G, ~3 W( k3 e2 R for (cnt = (maxsize / sizeof (long)) >> 1; cnt > 0; cnt >>= 1) {. |: \ q$ }+ I* Q7 H' E! }# y# s
addr = base + cnt; /* pointer arith! */( Y- ?9 m3 w. N9 b% s
sync ();: A( W5 q. L( {. h- G3 U2 L
save[i++] = *addr; # J' ]5 E9 t9 z! N sync (); / W9 U9 J$ T% A. K2 \8 o( T# r *addr = ~cnt;2 b( F" `6 E/ q1 O* V5 U
} : k4 J: F( S) u" H, x5 Q% }% T4 @; J1 S: B% r6 t8 z5 f
addr = base;' o! ~' g; o! ^( I' b* z
sync (); ) i P4 ?1 Z" ?% e save = *addr;; O4 {+ y- S! w/ N) j
sync ();' w( z; @# m& A1 I4 j) b" e* n" _
*addr = 0;7 p( _4 g2 R1 K3 ^5 V! \% ], z
) q- Z* K& H5 y5 ~) r) j7 E sync (); & _9 G9 |( G. \* \/ y& x4 i if ((val = *addr) != 0) {$ x, x: C; Q8 V! h) Z& z& ~
/* Restore the original data before leaving the function. 9 A/ B& f, [% [3 r7 z, Z */7 z/ o5 B( G) |7 \ \' b
sync ();6 U/ R% \7 j h& m! z
*addr = save;6 i! E( q3 D+ i- j1 b1 W
for (cnt = 1; cnt < maxsize / sizeof(long); cnt <<= 1) { 2 E7 c, }4 c; y i addr = base + cnt;+ J4 H" f1 A1 t+ E1 H; c
sync ();3 |5 ~4 h+ B) ?& [
*addr = save[--i]; * @ P2 N2 n# T7 I2 U6 e+ g } & q( h8 I: h0 X7 ~2 |! L return (0); + X9 @! [) E: [: g) x- B# l } 5 a9 c3 w+ x2 ` * y% k1 t% _1 U) @ for (cnt = 1; cnt < maxsize / sizeof (long); cnt <<= 1) {/ k& x+ q# Z& s3 b
addr = base + cnt; /* pointer arith! */ ) g# K9 \' y) h" S val = *addr; * D8 P: r% b6 r *addr = save[--i];7 t* v# i* Q6 G6 U: Q
if (val != ~cnt) { 7 R# R/ o* F* t' h0 U size = cnt * sizeof (long); # F4 [& S2 _/ T+ A /* Restore the original data before leaving the function. * x# A' ]# j6 Y% }+ ` Q */" |4 I$ E1 P: c1 t6 d1 i$ c
for (cnt <<= 1; cnt < maxsize / sizeof (long); cnt <<= 1) {" r u( e& R+ K
addr = base + cnt; 9 Z1 \1 t: Y5 c Q+ H1 K B' G *addr = save[--i];2 e w/ v) c, _6 Q+ I: r% i) s
}. b# |- T, o K1 G# o+ [
return (size);0 r4 A! P. @- i4 s. X' `) D
}( c7 r' v* N0 a2 v
} 3 E8 S4 l! `: M' Z! u4 x, H) `% O: e. O1 M/ e9 f: \+ d* ?
return (maxsize); 1 S! |/ z+ c6 e$ }! \} w( v l. U; D$ n" e, p) L
int dram_init(void)' n2 T( _: W3 Y [ s, `
{ ! @1 l4 y+ K- M6 g$ Y) n /* dram_init must store complete ramsize in gd->ram_size */ \8 C' W" x& B, J1 L: I
gd->ram_size = get_ram_size(! z" u; H4 Z8 p% A; L- }/ _& y+ i2 i
(void *)CONFIG_SYS_SDRAM_BASE, 4 F, k! [# q6 k, F CONFIG_MAX_RAM_BANK_SIZE); 4 ^7 G2 O9 a$ `7 ? return 0; 3 P: t+ s1 Z, c; T2 m7 V}0 ]% S' h4 A3 M8 R2 ~6 }# M" Y
$ e- F J8 y# j [ ]$ d# ~
- Z1 M7 N" ?6 D6 Z$ _
6 X& h- A# I: _: [) U4 o, B/ x Y m0 O7 f1 x" S
FLASH是通过检查FLASH内部的ID识别容量，希望对您有帮助！ " z- \( g. M) g4 X$ Y , T- r; I4 b; p' C! ?9 h# L; _: ^1 E, k/ p% K3 N, C& ]. r $ E4 w6 o$ b( P/ d# q% _6 [3 G