LLVM's IAS does not (and cannot easily) support the 'adrl'
pseudoinstruction, and ARM developers generally do not consider it
portable across assembler implementations either.
Instead, expand it into the two subtract instructions it would emit
anyway. An explanation of the math follows:
The .+8 and .+4 refer to the same memory location; this is because the
.+4 expression occurs in a subsequent instruction, 4 bytes after the
first. This memory location is the value of the PC register when it is
read by the first sub instruction. Thus, both inner parenthesized
expressions evaluate to the same result: PC's offset relative to
image_base. The subtract instructions then remove one byte each
(low, then high) of the total offset, thereby getting the absolute
address of image_base loaded in r0.
Signed-off-by: Sam Edwards <cfswo...@gmail.com>
---
arch/arm/lib/crt0_arm_efi.S | 3 ++-
1 file changed, 2 insertions(+), 1 deletion(-)
diff --git a/arch/arm/lib/crt0_arm_efi.S b/arch/arm/lib/crt0_arm_efi.S
index 91b0fe12c51..235b3a0c48f 100644
--- a/arch/arm/lib/crt0_arm_efi.S
+++ b/arch/arm/lib/crt0_arm_efi.S
@@ -149,7 +149,8 @@ _start:
adr r1, .L_DYNAMIC
ldr r0, [r1]
add r1, r0, r1
- adrl r0, image_base
+ sub r0, pc, #((.+8-image_base) & 0xff)
+ sub r0, r0, #((.+4-image_base) & 0xff00)
bl _relocate
teq r0, #0
bne 0f
--
2.45.2
