CODEWITHSUNDEEP
assemblygdbembeddedarmqemu
balavins
balavins

Reputation: 19

gdb cannot access memory for program running in qemu-arm emulator

I am trying to emulate a 'C' program by using qemu instruction level simulator.The 'C' program is compiled by issuing the following command

arm-none-linux-gnueabi-gcc -g ex_qsort.c -o ex_qsort_lin_work

I then start qemu with the following command

"qemu-arm -g 1234 -L /path/to/codesourcery/arm-none-linux-gnueabi/libc ./ex_qsort_lin_work"

Then I connect to the program using gdb. I am trying to access program memory location and change the assembly code. But when i try to access the memory I get the following error 

   (gdb) x 0x00008510
   0x8510 <main+76>:    0xe3530004
   (gdb) set *(0x8510) = 0xe3530002
   Cannot access memory at address 0x8510

I am not sure why this error occurs . Gdb does not give any other warning . When i start gdb I have the sysroot pointed to the arm library. However when i check for the shared libarary I get the following message 

   (gdb) info sharedlibrary 
   From        To          Syms Read   Shared Object Library
   0xf67d67d0  0xf67f0f58  Yes (*)     /path/to/codesourcery/arm-none-linux-gnueabi/libc/lib/ld-linux.so.3
   (*): Shared library is missing debugging information.

Not sure if this causes the problem. Statically linking the libraries also does not help My aim is to change the instruction at a given address

Upvotes: 0

Views: 2376

Answers (1)

Chris Dodd
Chris Dodd

Reputation: 126546

The issue is that the memory you are trying to modify is read-only. It looks like you're trying to modify code in the text segment of the executable, which is normally mapped read-only, so that is not unexpected. If you want to be able to modify it, it needs to be mapped as writable.

You can build an executable with a writable text segment (so it will load as writable by default) by linking with the -N flag -- use either -Xlinker -N or -Wl,-N on your gcc command line.

Upvotes: 1

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