On 26 February 2014 13:04, Gaurav Sharma <address@hidden> wrote:
Hi,
I have been trying to trace the for how address translation is done for any
load/store instructions. I was trying to emulate arm on an x86-64 machine.
However, i need some clarifications :
1. During the slow path, qemu uses helper functions to translate address.
2. This is done by calling the function itself during the execution.
3. The host instrn for the slow path is added at the end of the TB block. I
tried a sample code and got the following host instrn :
0x2aaade72d120: mov %r14,%rdi
0x2aaade72d123: xor %edx,%edx
0x2aaade72d125: lea -0x42(%rip),%rcx # 0x2aaade72d0ea
0x2aaade72d12c: mov $0x2afd98602c10,%r10
0x2aaade72d136: callq *%r10 // Call helper function
0x2aaade72d139: mov %eax,%ebp
0x2aaade72d13b: jmpq 0x2aaade72d0ea
3. How does it gets the address of the helper function :
call instruction is added by ' tcg_out_calli(s,
(uintptr_t)qemu_ld_helpers[opc & ~MO_SIGN]' line of code which fetches the
address of the helper function.
However from the assembly generated, the address is calculated before :
tcg_out_movi(s, TCG_TYPE_PTR, tcg_target_call_iarg_regs[3],
(uintptr_t)l->raddr)
This is just loading the 4th argument for the helper function into ECX
(which is the return address in generated code which corresponds to
the load we're going to do). It's not related to the address of the
helper function at all.
How is the address for the helper function calculated ?
You've just quoted the code that does it:
tcg_out_calli(s, (uintptr_t)qemu_ld_helpers[opc & ~MO_SIGN]' ...)
tcg_out_calli spots that the displacement is too big for a call insn
and emits the
0x2aaade72d12c: mov $0x2afd98602c10,%r10
0x2aaade72d136: callq *%r10 // Call helper function
thanks
-- PMM