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使用u-boot_2016_01启动mini2440(二)uboot内存分布

2016-04-04 20:54 302 查看
启动内存的分布主要是在board_init_f内划分的:

按照我的代码,分布如下:



代码log:

U-Boot 2016.01-g3401853-dirty (Apr 04 2016 - 08:06:26 -0400)

U-Boot code: 33A00000 -> 33A7BEF4  BSS: -> 33ACAAE4
CPUID: 32440001
FCLK:  405.600 MHz
HCLK:  101.400 MHz
PCLK:   50.700 MHz
DRAM:  Monitor len: 000CAAE4
Ram size: 04000000
Ram top: 34000000
TLB table from 33ff0000 to 33ff4000
LCD panel info: 320 x 240, 32 bit/pix
Reserving 300k for LCD Framebuffer at: 33fa5000
Reserving 810k for U-Boot at: 33eda000
Reserving 4096k for malloc() at: 33ada000
Reserving 80 Bytes for Board Info at: 33ad9fb0
Reserving 168 Bytes for Global Data at: 33ad9f08

RAM Configuration:
Bank #0: 30000000
DRAM:  64 MiB
New Stack Pointer is: 33ad9ee0
Relocation Offset is: 004da000
Relocating to 33eda000, new gd at 33ad9f08, sp at 33ad9ee0
WARNING: Caches not enabled
Flash: 0 Bytes
NAND:  256 MiB
MMC:   S3C MMC: 0
In:    serial
Out:   serial
Err:   serial
Net:   dm9000
MINI2440 #
MINI2440 # bdinfo
arch_number = 0x000007CF
boot_params = 0x30000100
DRAM bank   = 0x00000000
-> start    = 0x30000000
-> size     = 0x04000000
eth0name    = dm9000
ethaddr     = 08:00:3e:26:0a:5b
current eth = dm9000
ip_addr     = 192.168.1.10
baudrate    = 115200 bps
TLB addr    = 0x33FF0000
relocaddr   = 0x33EDA000
reloc off   = 0x004DA000
irq_sp      = 0x33AD9EF0
sp start    = 0x33AD9EE0
FB base     = 0x33FA5000
MINI2440 #
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