Ojs Ojs - 1 year ago 54
C Question

gdb can't access memory address error

here is my disas code:

0x0804844d <+0>: push %ebp
0x0804844e <+1>: mov %esp,%ebp
0x08048450 <+3>: and $0xfffffff0,%esp
0x08048453 <+6>: sub $0x20,%esp
0x08048456 <+9>: movl $0x8048540,(%esp)
0x0804845d <+16>: call 0x8048310 <[email protected]>
0x08048462 <+21>: lea 0x1c(%esp),%eax
0x08048466 <+25>: mov %eax,0x4(%esp)
0x0804846a <+29>: movl $0x8048555,(%esp)
0x08048471 <+36>: call 0x8048320 <[email protected]>
0x08048476 <+41>: mov 0x1c(%esp),%eax
0x0804847a <+45>: cmp $0x208c,%eax
0x0804847f <+50>: jne 0x804848f <main+66>
0x08048481 <+52>: movl $0x8048558,(%esp)
0x08048488 <+59>: call 0x8048310 <[email protected]>
0x0804848d <+64>: jmp 0x804849b <main+78>
=> 0x0804848f <+66>: movl $0x8048569,(%esp)
0x08048496 <+73>: call 0x8048310 <[email protected]>
0x0804849b <+78>: mov $0x0,%eax
0x080484a0 <+83>: leave
0x080484a1 <+84>: ret

what i'm tring to examine is $0x208c. When I type x/xw 0x208c it gives me back error which says Cannot access memory at address 0x208c. When i type Info registers and look at eax it says the value which i provided. So basically this program compares two values and depending on that prints something out.The problem is that this is homework from university and I have not got code. Hope you can help. Thank you.

Answer Source

When I type x/xw 0x208c it gives me back error which says Cannot access memory at address 0x208c

The disassembly for your program says that it does something like this:

puts("some string");
int i;
scanf("%d", &i);  // I don't know what the actual format string is.
                  // You can find out with x/s 0x8048555
if (i == 0x208c) { ... } else { ... }

In other words, the 0x208c is a value (8332) that your program has hard-coded in it, and is not a pointer. Therefore, GDB is entirely correct in telling you that if you interpret 0x208c as a pointer, that pointer does not point to readable memory.

i finally figured out to use print statement instead of x/xw

You appear to not understand the difference between print and examine commands. Consider this example:

int foo = 42;
int *pfoo = &foo;

With above, print pfoo will give you the address of foo, and x pfoo will give you the value stored at that address (i.e. the value of foo).

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