Hi all,
"ldrd Rm, Rn, <addr>" is an ARM instruction for loading two continuous words at
<addr> into two registers Rm, Rn. It can execute correctly but its dumping
information seems wrong. I wrote an simple assembly code to test:
.section .text, "ax"
global _start
_start:
mov r0, #0xff // r0 = 0xff
mov r1, #0x11 // r1 = 0x11
str r0, [sp], #4 // store 0x11 to <addr>, sp +=4
str r1, [sp], #4 // store 0x11 to <addr+4>, sp +=4
ldrD r2, r3, [sp, #-8] // r2 = value in <addr>, r3 = value in <addr+4>
mov r4, r2 // r4 = r2
mov r5, r3 // r5 = r3
# following is exiting.
exit:
mov r0, #0
mov r7, #1
svc 0
.end
Run it on SE mode and dump it with Exec flag:
1. 0x8054 : mov r0, #255 : IntAlu : D=0x00000000000000ff
2. 0x8058 : mov r1, #17 : IntAlu : D=0x0000000000000011
3. 0x805c.0 : str r0, [sp] #4 : MemWrite : D=0x00000000000000ff A=0xbeffff20
4. 0x805c.1 : addi_uop sp, sp, #4 : IntAlu : D=0x00000000beffff24
5. 0x8060.0 : str r1, [sp] #4 : MemWrite : D=0x0000000000000011 A=0xbeffff24
6. 0x8060.1 : addi_uop sp, sp, #4 : IntAlu : D=0x00000000beffff28
7. 0x8064 : ldrd r2, r3, [sp, #-8] : MemRead : D=0x0000000000000011 A=0xbeffff20
8. 0x8068 : mov r4, r2 : IntAlu : D=0x00000000000000ff
9. 0x806c : mov r5, r3 : IntAlu : D=0x0000000000000011
10. 0x8070 : mov r0, #0 : IntAlu : D=0x0000000000000000
11. 0x8074 : mov r7, #1 : IntAlu : D=0x0000000000000001
12. 0x8078 : svc : IntAlu :
In line 7 of dumped trace, should trace value "D=0x0000000000000011" be
"D=0x00000011000000ff"? (Assuming the configuration is little endian).
Thanks,
Meng
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