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From: "David Mc Kenna" <mckennad@esatclear.ie>
Reply-to: mckennad@esatclear.ie
To: Richard Earnshaw <rearnsha@arm.com>, mckennad@esatclear.ie, gdb@sources.redhat.com, insight@sources.redhat.com,   Richard.Earnshaw@arm.com
Cc: gdb@sources.redhat.com, insight@sources.redhat.com
Date: Wed, 03 Sep 2003 09:13:00 -0000
Subject: Re: ARM Simulator Bug? 
Message-id: <3f55b0d1.2580.0@esatclear.ie>
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Hi Richard,

>Try using an 
>
>	LDR r0, =__thumb
>

This command places the correct value into R0 for the BX command, but the bug
within the simulator is still present. 

Take the following code as an example.

0000005c <TST_start>:
  5c:   e3a0da12        mov     sp, #73728      ; 0x12000
  60:   e59f00a4        ldr     r0, [pc, #164]  ; 10c <IRQ_Interrupt+0x50>
  64:   e12fff10        bx      r0

00000068 <__start_of_thumb>:

void    IRQ_Interrupt(void);

int main()
{
  68:   b580            push    {r7, lr}
  6a:   466f            mov     r7, sp
  6c:   b082            sub     sp, #8

If I place a breakpoint at 0x6c and run the code gdb loses itself, i.e. if I
interrupt it with ctrl-c it says the PC is 0x0580fa70 or some thing similar.
Again, if I place a breakpoint at 0x5c, I can single step successfully. If I
single step to 0x68 and place a breakpoint at 0x6c and continue, gdb loses itself
again.

This problem only occurs when I attempt to use thumb. If I remove the bx command
and compile my C file in ARM mode the simulator works perfectly.

Is it possible to see which commands the simulator is executing? 

Thanks,
Dave


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