嵌入式开发者社区
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
) W8 c5 T& d( _# Y" ^0 X4 |2 O4 M! @ f
核心板2:DDR2 256M Byte NAND FLASH 8G bit
& f$ z. |$ O) i. k2 J
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?
4 D& {; }, O1 I4 }
7 }; o# o) D+ ^* m. f8 e" |
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?
, u( j; w7 V2 d% ?) A
6 U5 b. `* k+ U' r2 ^
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:
1 E8 S, C( c3 ~& D" Y e5 X
/*
! Y& I4 I8 T7 j8 N
* Check memory range for valid RAM. A simple memory test determines
% S" n# C$ ~ `' i! R. A4 [+ e
* the actually available RAM size between addresses `base' and
* `. K' ^5 E0 }; \8 b$ I
* `base + maxsize'.
4 ]% i1 U1 {& L- x9 @8 [
*/
9 b: G! g1 L1 E- S
long get_ram_size(long *base, long maxsize)
/ k$ ^( H5 ~+ ~5 |: d7 {- F: M
{
0 T0 k @# t5 L& S" I
volatile long *addr;
4 `, {% y. o$ ^3 n) V% G9 b& t
long save[32];
/ K8 N7 t! S: I T8 J
long cnt;
+ S' h0 _8 s( ~
long val;
/ p) S% _8 r2 U- S8 e6 U
long size;
; O- `& k# T X, J
int i = 0;
$ ~' w1 a- G9 v, {* [$ g& p
" d4 d8 q$ M. }1 H9 @
for (cnt = (maxsize / sizeof (long)) >> 1; cnt > 0; cnt >>= 1) {
" C3 q$ U# O2 B# C1 w: O$ k! M
addr = base + cnt; /* pointer arith! */
4 t, I, w+ F9 `* N0 B; n
sync ();
1 W4 c1 Z8 \- j; b7 R
save[i++] = *addr;
/ c' B3 \4 M( @& ^
sync ();
7 _* H9 f; ^/ s- L8 K8 l1 a: w8 N1 j
*addr = ~cnt;
8 i" U' t. Q+ s# h( y
}
! s, @/ f( c+ I* {: z+ m
! l( I! U: S6 K! \. q* F$ ?$ x
addr = base;
/ ]# a; O# Z$ ?, M7 E3 A
sync ();
1 B, t w& H% C% l. F3 S3 m
save
= *addr;
) y4 s0 X; }; y2 M: o
sync ();
6 ^4 P" J1 e" e! y; y$ z2 N! |5 g
*addr = 0;
J# G0 }# B" L8 ^
1 [3 L* ~/ b7 ~& s u" n
sync ();
2 t* c. \, T) p
if ((val = *addr) != 0) {
1 R1 T0 O) l. I R) b7 [1 m! G
/* Restore the original data before leaving the function.
8 p8 [6 U$ I9 n1 M& J
*/
; l! Y1 u4 N! z \4 b. m
sync ();
. C/ C9 l$ ~# \4 y7 ?
*addr = save
;
) G% Y: }4 J# T0 E7 C1 I1 R
for (cnt = 1; cnt < maxsize / sizeof(long); cnt <<= 1) {
& C U+ W8 c1 m! x! \- i; W8 r0 L
addr = base + cnt;
# Q k! L! }) o4 o6 j: `+ c5 I
sync ();
% l/ {8 F, F7 j( s6 V
*addr = save[--i];
( o% g6 F1 w2 ^9 z( X# h7 ?! K8 i4 `
}
/ Z$ y- c e8 ]" m8 ]% {
return (0);
% P! C( T4 j: ^" W3 [& W
}
L6 f7 s, ~' W6 A# m6 h! k
`# F& K$ L3 H, m/ Z" E U
for (cnt = 1; cnt < maxsize / sizeof (long); cnt <<= 1) {
' H) D- K& X5 p8 g7 N/ q$ `4 s
addr = base + cnt; /* pointer arith! */
, c" h$ W4 E4 J6 f
val = *addr;
v5 w3 X3 [, l5 _% {
*addr = save[--i];
. e# |% O% m0 T1 |9 N
if (val != ~cnt) {
6 i7 b! C9 x; q# r
size = cnt * sizeof (long);
7 c8 N1 f( N. x3 i3 n. t
/* Restore the original data before leaving the function.
) J/ x2 o0 ~2 X7 S" l9 N+ c$ d
*/
& T8 K" K( V4 E. L7 Z1 b+ A
for (cnt <<= 1; cnt < maxsize / sizeof (long); cnt <<= 1) {
& P8 ~" R2 C, s" o
addr = base + cnt;
7 g- ?5 I9 f8 Q9 Q. \, m6 @' Q
*addr = save[--i];
! _& a* |3 p1 ~1 Y' w& ^8 e& @
}
( }5 F; ?9 U5 M) Z! k' [& |. W; a
return (size);
; F4 @7 } |7 q" t3 W$ j3 P
}
6 u, d, L$ [) B# }' N R8 D' g
}
* j+ G1 ~) i& f2 h6 b! D
- v4 L6 \' p/ {+ |) {% r
return (maxsize);
# R+ V# g" H, B O, C2 R0 b; k
}
& F4 R3 `3 c! ~& n y2 `
int dram_init(void)
5 {# a- P* V- d& C
{
( }; M- R* [. ?
/* dram_init must store complete ramsize in gd->ram_size */
" M9 l2 i e+ N" ?( {# T# R" l) i
gd->ram_size = get_ram_size(
: k( t$ M+ L+ a- e0 \0 f- A
(void *)CONFIG_SYS_SDRAM_BASE,
) R8 X* k1 N! [; @3 s* I+ \, \, S, i
CONFIG_MAX_RAM_BANK_SIZE);
0 U( P4 K) b8 s- U
return 0;
& C: E ]3 O; N, {3 ~! S. r! B
}
! z2 o& L* {& z( h5 z. Y
! b; b* w" f$ S k6 {$ V/ `0 @
. z9 F0 N# X/ w
b+ |, z" a5 D$ ]
3 N" J5 n n+ A0 D/ V7 E4 W. D
FLASH是通过检查FLASH内部的ID识别容量,希望对您有帮助!
6 C( `, A! x) y/ F; N1 N: Z
' f [4 i9 _' q- d2 F' Y
+ F) z8 h: H- ^+ |2 |3 z
B! Q9 X: l# N# k+ u
欢迎光临 嵌入式开发者社区 (https://www.51ele.net/)
Powered by Discuz! X3.4