在uboot的920T中有如下一段代码: mov r0, #0 ldr r1, uboot_ram_base_addr mov r2, #0x400 @ compare first 4-K bytes 1001: ldr r3, [r0], #4 ldr r4, [r1], #4 teq r3, r4 bne 1002f @ not matched subs r2, r2, #4 beq 1003f b 1001b
1002: ldr r0, =GPFDAT ldr r1, =0x30 str r1,[r0]
1: b 1b
1003:
ldr r0, =GPFDAT ldr r1, =0x40 str r1,[r0]
@ jump to ram ldr r0, uboot_ram_base_addr add pc, r10, r0 /* fake ^^; return here. */
copy_myself_failed: 1: b 1b
这里面为什么每个跳转(比如1001,1002,1003)后面都带一个字母呢?如(1001f,1002f,1003b)等,理解为二进制或者十六进制似乎不妥。请教各位,有没有更好的解释? |