Cant access a specific memory address. on C program
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2
down vote
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I am working on the book "HACKING Art Of Exploitation " exercise Convert2.c page 61.
Here's my code. Below is my question.
#include <stdio.h>
void usage(char *program_name) {
printf("Usage: %s <message> <# of times to repeat>n", program_name);
exit(1);
}
int main(int argc, char *argv) {
int i, count;
// if(argc < 3) //if fewer than 3 arguments is used
// usage(argv[0]); // display usage message and exit
count = atoi(argv[2]); //convert the second arg into an interger
printf("Repeating %d timesn", count);
for(i=0; i < count; i++)
printf("%3d - %sn", i, argv[1]); // print the first arg
}
GDB OUTPUT...
➜ git:(master) ✗ 👽 gdb -q a.out
Reading symbols from a.out...done.
(gdb) run test
Starting program: /home/fruitdealer/clones/C_zombie/hacking/a.out test
Program received signal SIGSEGV, Segmentation fault.
__GI_____strtol_l_internal (nptr=0x0, endptr=endptr@entry=0x0, base=base@entry=10, group=group@entry=0,
loc=0x7ffff7dd0560 <_nl_global_locale>) at ../stdlib/strtol_l.c:292
292 ../stdlib/strtol_l.c: No such file or directory.
(gdb) break main
Breakpoint 1 at 0x555555554707: file convert.c, line 14.
(gdb) where
#0 __GI_____strtol_l_internal (nptr=0x0, endptr=endptr@entry=0x0, base=base@entry=10, group=group@entry=0,
loc=0x7ffff7dd0560 <_nl_global_locale>) at ../stdlib/strtol_l.c:292
#1 0x00007ffff7a29122 in __strtol (nptr=<optimized out>, endptr=endptr@entry=0x0, base=base@entry=10)
at ../stdlib/strtol.c:106
#2 0x00007ffff7a24690 in atoi (nptr=<optimized out>) at atoi.c:27
#3 0x000055555555471f in main (argc=2, argv=0x7fffffffdeb8) at convert.c:14
(gdb) run test
The program being debugged has been started already.
Start it from the beginning? (y or n) y
Starting program: /home/fruitdealer/clones/C_zombie/hacking/a.out test
Breakpoint 1, main (argc=2, argv=0x7fffffffdeb8) at convert.c:14
14 count = atoi(argv[2]); //convert the second arg into an interger
(gdb) cont
Continuing.
Program received signal SIGSEGV, Segmentation fault.
__GI_____strtol_l_internal (nptr=0x0, endptr=endptr@entry=0x0, base=base@entry=10, group=group@entry=0,
loc=0x7ffff7dd0560 <_nl_global_locale>) at ../stdlib/strtol_l.c:292
292 ../stdlib/strtol_l.c: No such file or directory.
(gdb) x/3xw 0x7fffffffdeb8
0x7fffffffdeb8: 0xffffe220 0x00007fff 0xffffe250
(gdb) x/s 0xffffe220
0xffffe220: <error: Cannot access memory at address 0xffffe220>
(gdb) x/s 0xffffe250
0xffffe250: <error: Cannot access memory at address 0xffffe250>
(gdb) x/sw 0xffffe250
0xffffe250: <error: Cannot access memory at address 0xffffe250>
(gdb)
I posted all of gdb output because i wasn't sure how much of it you would need. My problem lies at the bottom of my GDB output when i run "x/s" on gdb and get the <error: Cannot access memory at address 0xffffe250>
error.
On the book Jon Erickson is able to access 0xffffe220
and 0x00007fff
and then he has an error on 0xffffe250
this part of memory.
What am i missing?
Why can't i access any of the three addresses in 0x7fffffffdeb8
?
c linux gcc memory gdb
add a comment |
up vote
2
down vote
favorite
I am working on the book "HACKING Art Of Exploitation " exercise Convert2.c page 61.
Here's my code. Below is my question.
#include <stdio.h>
void usage(char *program_name) {
printf("Usage: %s <message> <# of times to repeat>n", program_name);
exit(1);
}
int main(int argc, char *argv) {
int i, count;
// if(argc < 3) //if fewer than 3 arguments is used
// usage(argv[0]); // display usage message and exit
count = atoi(argv[2]); //convert the second arg into an interger
printf("Repeating %d timesn", count);
for(i=0; i < count; i++)
printf("%3d - %sn", i, argv[1]); // print the first arg
}
GDB OUTPUT...
➜ git:(master) ✗ 👽 gdb -q a.out
Reading symbols from a.out...done.
(gdb) run test
Starting program: /home/fruitdealer/clones/C_zombie/hacking/a.out test
Program received signal SIGSEGV, Segmentation fault.
__GI_____strtol_l_internal (nptr=0x0, endptr=endptr@entry=0x0, base=base@entry=10, group=group@entry=0,
loc=0x7ffff7dd0560 <_nl_global_locale>) at ../stdlib/strtol_l.c:292
292 ../stdlib/strtol_l.c: No such file or directory.
(gdb) break main
Breakpoint 1 at 0x555555554707: file convert.c, line 14.
(gdb) where
#0 __GI_____strtol_l_internal (nptr=0x0, endptr=endptr@entry=0x0, base=base@entry=10, group=group@entry=0,
loc=0x7ffff7dd0560 <_nl_global_locale>) at ../stdlib/strtol_l.c:292
#1 0x00007ffff7a29122 in __strtol (nptr=<optimized out>, endptr=endptr@entry=0x0, base=base@entry=10)
at ../stdlib/strtol.c:106
#2 0x00007ffff7a24690 in atoi (nptr=<optimized out>) at atoi.c:27
#3 0x000055555555471f in main (argc=2, argv=0x7fffffffdeb8) at convert.c:14
(gdb) run test
The program being debugged has been started already.
Start it from the beginning? (y or n) y
Starting program: /home/fruitdealer/clones/C_zombie/hacking/a.out test
Breakpoint 1, main (argc=2, argv=0x7fffffffdeb8) at convert.c:14
14 count = atoi(argv[2]); //convert the second arg into an interger
(gdb) cont
Continuing.
Program received signal SIGSEGV, Segmentation fault.
__GI_____strtol_l_internal (nptr=0x0, endptr=endptr@entry=0x0, base=base@entry=10, group=group@entry=0,
loc=0x7ffff7dd0560 <_nl_global_locale>) at ../stdlib/strtol_l.c:292
292 ../stdlib/strtol_l.c: No such file or directory.
(gdb) x/3xw 0x7fffffffdeb8
0x7fffffffdeb8: 0xffffe220 0x00007fff 0xffffe250
(gdb) x/s 0xffffe220
0xffffe220: <error: Cannot access memory at address 0xffffe220>
(gdb) x/s 0xffffe250
0xffffe250: <error: Cannot access memory at address 0xffffe250>
(gdb) x/sw 0xffffe250
0xffffe250: <error: Cannot access memory at address 0xffffe250>
(gdb)
I posted all of gdb output because i wasn't sure how much of it you would need. My problem lies at the bottom of my GDB output when i run "x/s" on gdb and get the <error: Cannot access memory at address 0xffffe250>
error.
On the book Jon Erickson is able to access 0xffffe220
and 0x00007fff
and then he has an error on 0xffffe250
this part of memory.
What am i missing?
Why can't i access any of the three addresses in 0x7fffffffdeb8
?
c linux gcc memory gdb
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am working on the book "HACKING Art Of Exploitation " exercise Convert2.c page 61.
Here's my code. Below is my question.
#include <stdio.h>
void usage(char *program_name) {
printf("Usage: %s <message> <# of times to repeat>n", program_name);
exit(1);
}
int main(int argc, char *argv) {
int i, count;
// if(argc < 3) //if fewer than 3 arguments is used
// usage(argv[0]); // display usage message and exit
count = atoi(argv[2]); //convert the second arg into an interger
printf("Repeating %d timesn", count);
for(i=0; i < count; i++)
printf("%3d - %sn", i, argv[1]); // print the first arg
}
GDB OUTPUT...
➜ git:(master) ✗ 👽 gdb -q a.out
Reading symbols from a.out...done.
(gdb) run test
Starting program: /home/fruitdealer/clones/C_zombie/hacking/a.out test
Program received signal SIGSEGV, Segmentation fault.
__GI_____strtol_l_internal (nptr=0x0, endptr=endptr@entry=0x0, base=base@entry=10, group=group@entry=0,
loc=0x7ffff7dd0560 <_nl_global_locale>) at ../stdlib/strtol_l.c:292
292 ../stdlib/strtol_l.c: No such file or directory.
(gdb) break main
Breakpoint 1 at 0x555555554707: file convert.c, line 14.
(gdb) where
#0 __GI_____strtol_l_internal (nptr=0x0, endptr=endptr@entry=0x0, base=base@entry=10, group=group@entry=0,
loc=0x7ffff7dd0560 <_nl_global_locale>) at ../stdlib/strtol_l.c:292
#1 0x00007ffff7a29122 in __strtol (nptr=<optimized out>, endptr=endptr@entry=0x0, base=base@entry=10)
at ../stdlib/strtol.c:106
#2 0x00007ffff7a24690 in atoi (nptr=<optimized out>) at atoi.c:27
#3 0x000055555555471f in main (argc=2, argv=0x7fffffffdeb8) at convert.c:14
(gdb) run test
The program being debugged has been started already.
Start it from the beginning? (y or n) y
Starting program: /home/fruitdealer/clones/C_zombie/hacking/a.out test
Breakpoint 1, main (argc=2, argv=0x7fffffffdeb8) at convert.c:14
14 count = atoi(argv[2]); //convert the second arg into an interger
(gdb) cont
Continuing.
Program received signal SIGSEGV, Segmentation fault.
__GI_____strtol_l_internal (nptr=0x0, endptr=endptr@entry=0x0, base=base@entry=10, group=group@entry=0,
loc=0x7ffff7dd0560 <_nl_global_locale>) at ../stdlib/strtol_l.c:292
292 ../stdlib/strtol_l.c: No such file or directory.
(gdb) x/3xw 0x7fffffffdeb8
0x7fffffffdeb8: 0xffffe220 0x00007fff 0xffffe250
(gdb) x/s 0xffffe220
0xffffe220: <error: Cannot access memory at address 0xffffe220>
(gdb) x/s 0xffffe250
0xffffe250: <error: Cannot access memory at address 0xffffe250>
(gdb) x/sw 0xffffe250
0xffffe250: <error: Cannot access memory at address 0xffffe250>
(gdb)
I posted all of gdb output because i wasn't sure how much of it you would need. My problem lies at the bottom of my GDB output when i run "x/s" on gdb and get the <error: Cannot access memory at address 0xffffe250>
error.
On the book Jon Erickson is able to access 0xffffe220
and 0x00007fff
and then he has an error on 0xffffe250
this part of memory.
What am i missing?
Why can't i access any of the three addresses in 0x7fffffffdeb8
?
c linux gcc memory gdb
I am working on the book "HACKING Art Of Exploitation " exercise Convert2.c page 61.
Here's my code. Below is my question.
#include <stdio.h>
void usage(char *program_name) {
printf("Usage: %s <message> <# of times to repeat>n", program_name);
exit(1);
}
int main(int argc, char *argv) {
int i, count;
// if(argc < 3) //if fewer than 3 arguments is used
// usage(argv[0]); // display usage message and exit
count = atoi(argv[2]); //convert the second arg into an interger
printf("Repeating %d timesn", count);
for(i=0; i < count; i++)
printf("%3d - %sn", i, argv[1]); // print the first arg
}
GDB OUTPUT...
➜ git:(master) ✗ 👽 gdb -q a.out
Reading symbols from a.out...done.
(gdb) run test
Starting program: /home/fruitdealer/clones/C_zombie/hacking/a.out test
Program received signal SIGSEGV, Segmentation fault.
__GI_____strtol_l_internal (nptr=0x0, endptr=endptr@entry=0x0, base=base@entry=10, group=group@entry=0,
loc=0x7ffff7dd0560 <_nl_global_locale>) at ../stdlib/strtol_l.c:292
292 ../stdlib/strtol_l.c: No such file or directory.
(gdb) break main
Breakpoint 1 at 0x555555554707: file convert.c, line 14.
(gdb) where
#0 __GI_____strtol_l_internal (nptr=0x0, endptr=endptr@entry=0x0, base=base@entry=10, group=group@entry=0,
loc=0x7ffff7dd0560 <_nl_global_locale>) at ../stdlib/strtol_l.c:292
#1 0x00007ffff7a29122 in __strtol (nptr=<optimized out>, endptr=endptr@entry=0x0, base=base@entry=10)
at ../stdlib/strtol.c:106
#2 0x00007ffff7a24690 in atoi (nptr=<optimized out>) at atoi.c:27
#3 0x000055555555471f in main (argc=2, argv=0x7fffffffdeb8) at convert.c:14
(gdb) run test
The program being debugged has been started already.
Start it from the beginning? (y or n) y
Starting program: /home/fruitdealer/clones/C_zombie/hacking/a.out test
Breakpoint 1, main (argc=2, argv=0x7fffffffdeb8) at convert.c:14
14 count = atoi(argv[2]); //convert the second arg into an interger
(gdb) cont
Continuing.
Program received signal SIGSEGV, Segmentation fault.
__GI_____strtol_l_internal (nptr=0x0, endptr=endptr@entry=0x0, base=base@entry=10, group=group@entry=0,
loc=0x7ffff7dd0560 <_nl_global_locale>) at ../stdlib/strtol_l.c:292
292 ../stdlib/strtol_l.c: No such file or directory.
(gdb) x/3xw 0x7fffffffdeb8
0x7fffffffdeb8: 0xffffe220 0x00007fff 0xffffe250
(gdb) x/s 0xffffe220
0xffffe220: <error: Cannot access memory at address 0xffffe220>
(gdb) x/s 0xffffe250
0xffffe250: <error: Cannot access memory at address 0xffffe250>
(gdb) x/sw 0xffffe250
0xffffe250: <error: Cannot access memory at address 0xffffe250>
(gdb)
I posted all of gdb output because i wasn't sure how much of it you would need. My problem lies at the bottom of my GDB output when i run "x/s" on gdb and get the <error: Cannot access memory at address 0xffffe250>
error.
On the book Jon Erickson is able to access 0xffffe220
and 0x00007fff
and then he has an error on 0xffffe250
this part of memory.
What am i missing?
Why can't i access any of the three addresses in 0x7fffffffdeb8
?
c linux gcc memory gdb
c linux gcc memory gdb
edited Nov 10 at 1:01
asked Nov 10 at 0:46
Joel alexis
154
154
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
The first half of the address is cut off. If you notice, it takes 8 bytes to store the addresses because you are on a 64 bit machine, not 32. You are trying to access a truncated address.
Rather than three addresses at 0x7fffffffdeb8
, you are looking at one and a half. Try accessing a byte that starts with 0x00007fff
...
2
In other words, don't usex/3xw
, usex/3gx
instead.
– Employed Russian
Nov 10 at 1:16
Ok i see. Thank you. I will usex/3gx
the "g" for giant. Can you tell me how you are able to tell this is a 64-bit system just by looking at my gdb output? Blessings.
– Joel alexis
Nov 10 at 15:01
1
@Joelalexis you can tell by looking at the addresses the GDB outputs. On a 32-bit system, the memory address range is [0 to ((2^32)-1)] which is equal to [0x00000000 - 0xffffffff]. If you see an address that is outside this range in the GDB output (such as "loc=0x7ffff7dd0560"), that indicates you are on a 64-bit system. Alternatively you can examine the stack, as you are already doing.
– jahneff
Nov 10 at 19:38
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The first half of the address is cut off. If you notice, it takes 8 bytes to store the addresses because you are on a 64 bit machine, not 32. You are trying to access a truncated address.
Rather than three addresses at 0x7fffffffdeb8
, you are looking at one and a half. Try accessing a byte that starts with 0x00007fff
...
2
In other words, don't usex/3xw
, usex/3gx
instead.
– Employed Russian
Nov 10 at 1:16
Ok i see. Thank you. I will usex/3gx
the "g" for giant. Can you tell me how you are able to tell this is a 64-bit system just by looking at my gdb output? Blessings.
– Joel alexis
Nov 10 at 15:01
1
@Joelalexis you can tell by looking at the addresses the GDB outputs. On a 32-bit system, the memory address range is [0 to ((2^32)-1)] which is equal to [0x00000000 - 0xffffffff]. If you see an address that is outside this range in the GDB output (such as "loc=0x7ffff7dd0560"), that indicates you are on a 64-bit system. Alternatively you can examine the stack, as you are already doing.
– jahneff
Nov 10 at 19:38
add a comment |
up vote
3
down vote
The first half of the address is cut off. If you notice, it takes 8 bytes to store the addresses because you are on a 64 bit machine, not 32. You are trying to access a truncated address.
Rather than three addresses at 0x7fffffffdeb8
, you are looking at one and a half. Try accessing a byte that starts with 0x00007fff
...
2
In other words, don't usex/3xw
, usex/3gx
instead.
– Employed Russian
Nov 10 at 1:16
Ok i see. Thank you. I will usex/3gx
the "g" for giant. Can you tell me how you are able to tell this is a 64-bit system just by looking at my gdb output? Blessings.
– Joel alexis
Nov 10 at 15:01
1
@Joelalexis you can tell by looking at the addresses the GDB outputs. On a 32-bit system, the memory address range is [0 to ((2^32)-1)] which is equal to [0x00000000 - 0xffffffff]. If you see an address that is outside this range in the GDB output (such as "loc=0x7ffff7dd0560"), that indicates you are on a 64-bit system. Alternatively you can examine the stack, as you are already doing.
– jahneff
Nov 10 at 19:38
add a comment |
up vote
3
down vote
up vote
3
down vote
The first half of the address is cut off. If you notice, it takes 8 bytes to store the addresses because you are on a 64 bit machine, not 32. You are trying to access a truncated address.
Rather than three addresses at 0x7fffffffdeb8
, you are looking at one and a half. Try accessing a byte that starts with 0x00007fff
...
The first half of the address is cut off. If you notice, it takes 8 bytes to store the addresses because you are on a 64 bit machine, not 32. You are trying to access a truncated address.
Rather than three addresses at 0x7fffffffdeb8
, you are looking at one and a half. Try accessing a byte that starts with 0x00007fff
...
edited Nov 10 at 1:18
answered Nov 10 at 1:07
jahneff
433
433
2
In other words, don't usex/3xw
, usex/3gx
instead.
– Employed Russian
Nov 10 at 1:16
Ok i see. Thank you. I will usex/3gx
the "g" for giant. Can you tell me how you are able to tell this is a 64-bit system just by looking at my gdb output? Blessings.
– Joel alexis
Nov 10 at 15:01
1
@Joelalexis you can tell by looking at the addresses the GDB outputs. On a 32-bit system, the memory address range is [0 to ((2^32)-1)] which is equal to [0x00000000 - 0xffffffff]. If you see an address that is outside this range in the GDB output (such as "loc=0x7ffff7dd0560"), that indicates you are on a 64-bit system. Alternatively you can examine the stack, as you are already doing.
– jahneff
Nov 10 at 19:38
add a comment |
2
In other words, don't usex/3xw
, usex/3gx
instead.
– Employed Russian
Nov 10 at 1:16
Ok i see. Thank you. I will usex/3gx
the "g" for giant. Can you tell me how you are able to tell this is a 64-bit system just by looking at my gdb output? Blessings.
– Joel alexis
Nov 10 at 15:01
1
@Joelalexis you can tell by looking at the addresses the GDB outputs. On a 32-bit system, the memory address range is [0 to ((2^32)-1)] which is equal to [0x00000000 - 0xffffffff]. If you see an address that is outside this range in the GDB output (such as "loc=0x7ffff7dd0560"), that indicates you are on a 64-bit system. Alternatively you can examine the stack, as you are already doing.
– jahneff
Nov 10 at 19:38
2
2
In other words, don't use
x/3xw
, use x/3gx
instead.– Employed Russian
Nov 10 at 1:16
In other words, don't use
x/3xw
, use x/3gx
instead.– Employed Russian
Nov 10 at 1:16
Ok i see. Thank you. I will use
x/3gx
the "g" for giant. Can you tell me how you are able to tell this is a 64-bit system just by looking at my gdb output? Blessings.– Joel alexis
Nov 10 at 15:01
Ok i see. Thank you. I will use
x/3gx
the "g" for giant. Can you tell me how you are able to tell this is a 64-bit system just by looking at my gdb output? Blessings.– Joel alexis
Nov 10 at 15:01
1
1
@Joelalexis you can tell by looking at the addresses the GDB outputs. On a 32-bit system, the memory address range is [0 to ((2^32)-1)] which is equal to [0x00000000 - 0xffffffff]. If you see an address that is outside this range in the GDB output (such as "loc=0x7ffff7dd0560"), that indicates you are on a 64-bit system. Alternatively you can examine the stack, as you are already doing.
– jahneff
Nov 10 at 19:38
@Joelalexis you can tell by looking at the addresses the GDB outputs. On a 32-bit system, the memory address range is [0 to ((2^32)-1)] which is equal to [0x00000000 - 0xffffffff]. If you see an address that is outside this range in the GDB output (such as "loc=0x7ffff7dd0560"), that indicates you are on a 64-bit system. Alternatively you can examine the stack, as you are already doing.
– jahneff
Nov 10 at 19:38
add a comment |
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