How to modify memory contents using GDB?

2019-01-12 21:45发布

问题:

I know that we can use several commands to access and read memory: for example, print, p, x...

But how can I change the contents of memory at any specific location (while debugging in GDB)?

回答1:

The easiest is setting a program variable (see GDB: assignment):

(gdb) l
6       {
7           int i;
8           struct file *f, *ftmp;
9
(gdb) set variable i = 10
(gdb) p i
$1 = 10

Or you can just update arbitrary (writable) location by address:

(gdb) set {int}0x83040 = 4

There's more. Read the manual.



回答2:

As Nikolai has said you can use the gdb 'set' command to change the value of a variable.

You can also use the 'set' command to change memory locations. eg. Expanding on Nikolai's example:

(gdb) l
6       {
7           int i;
8           struct file *f, *ftmp;
9
(gdb) set variable i = 10
(gdb) p i
$1 = 10

(gdb) p &i
$2 = (int *) 0xbfbb0000
(gdb) set *((int *) 0xbfbb0000) = 20
(gdb) p i
$3 = 20

This should work for any valid pointer, and can be cast to any appropriate data type.



回答3:

Expanding on the answers provided here.

You can just do set idx = 1 to set a variable, but that syntax is not recommended because the variable name may clash with a set sub-command. As an example set w=1 would not be valid.

This means that you should prefer the syntax: set variable idx = 1 or set var idx = 1.

Last but not least, you can just use your trusty old print command, since it evaluates an expression. The only difference being that he also prints the result of the expression.

(gdb) p idx = 1
$1 = 1

You can read more about gdb here.



标签: c linux memory gdb