[U-Boot] A bl command question about smdk2410 in u-boot-2010.06

[Jianchun Zhou]

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!


-- 
Best Regards