Hi, folks:

I am new to U-Boot, and now learning the code of it.

I have configured u-boot for smdk2410 board, and I really got a binary file named u-boot.

And I disassembled it, I got this fragment of code(generated from source code file start.S):

..........

33f80050 <start_code>: 33f80050: e10f0000 mrs r0, CPSR 33f80054: e3c0001f bic r0, r0, #31 ; 0x1f 33f80058: e38000d3 orr r0, r0, #211 ; 0xd3 33f8005c: e129f000 msr CPSR_fc, r0 33f80060: eb0000e6 bl 33f80400 <__coloured_LED_init> 33f80064: eb0000e6 bl 33f80404 <__red_LED_on> 33f80068: e3a00453 mov r0, #1392508928 ; 0x53000000 33f8006c: e3a01000 mov r1, #0 ; 0x0 33f80070: e5801000 str r1, [r0] 33f80074: e3e01000 mvn r1, #0 ; 0x0 33f80078: e59f0360 ldr r0, [pc, #864] ; 33f803e0 <fiq+0x60> 33f8007c: e5801000 str r1, [r0] 33f80080: e59f135c ldr r1, [pc, #860] ; 33f803e4 <fiq+0x64> 33f80084: e59f035c ldr r0, [pc, #860] ; 33f803e8 <fiq+0x68> 33f80088: e5801000 str r1, [r0] 33f8008c: e59f0358 ldr r0, [pc, #856] ; 33f803ec <fiq+0x6c> 33f80090: e3a01003 mov r1, #3 ; 0x3 33f80094: e5801000 str r1, [r0] 33f80098: eb000019 bl 33f80104 <cpu_init_crit>

33f8009c <relocate>: 33f8009c: e24f00a4 sub r0, pc, #164 ; 0xa4 33f800a0: e51f1068 ldr r1, [pc, #-104] ; 33f80040 <_TEXT_BASE> .............

here I am wondering in the code that is highlighted with red color what will happen when we jump to addr 33f80104 (in SDRAM addr space)

if you know at this time the code is executing SDRAM memory is not initialized yet?

Anybody can help me out? Thanks in advance!