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Re: disassembly view
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Re: disassembly view


  • Subject: Re: disassembly view
  • From: Jonas Maebe <email@hidden>
  • Date: Mon, 1 Aug 2005 08:19:08 +0200


On 1 aug 2005, at 02:21, email@hidden wrote:


When debugging programs compiled with the Free Pascal Compiler, the disassembly view in XCode doesn't work (breakpoints work fine). The reason is that XCode sends the following command to gdb in all cases (no matter where you are in the code):

# Executing Sequence: <PBXGDB_DisassembleSequence: 0x558c7e0> t=3.714260 Tepoch=1122839425.887780
-> 1285-data-disassemble -s 18446744073709551360 -e 256 -- 1


My question: how does XCode determine that it will send the address "18446744073709551360" to gdb? This is obviously bogus, but I cannot find any indication in the XCode-gdb communication log where XCode gets this value from.


Xcode is trying to disassemble addresses around the pc.


How does it obtain the pc? gdb prints the correct value with both "info registers" and "p $pc". These are 32 bit binaries, so I'm not sure where the 64bit address comes from.


Thanks,


Jonas

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References: 
 >disassembly view (From: Jonas Maebe <email@hidden>)
 >Re: disassembly view (From: email@hidden)

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