We are dealing with C here. I'm just had this idea, wondering if it is possible to access the point in memory where a function is stored, say foo
and copying the contents of the function to another point in memory. Specifically, I'm trying to get the following to work:
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
void foo(){
printf("Hello World");
}
int main(){
void (*bar)(void) = malloc(sizeof foo);
memcpy(&bar, &foo, sizeof foo);
bar();
return 0;
}
But running it gives a bus error: Bus error: 10
. I'm trying to copy over the contents of function foo
into a space of memory bar
and then executing the newly created function bar
.
This is for no other reason than to see if such a thing is possible, to reveal the intricacies of the C language. I'm not thinking about what practical uses this has.
I'm looking for guidance getting this to work, or otherwise to be told, with a reason, why this won't work
EDIT Looking at some of the answers and learning about read, write, and executable memory, it just dawned upon me that it would be possible to create functions on the fly in C by writing to executable memory.
With standard C, what you try to do is implementation defined behaviour and won't work portably. On a given platform, you might be able to make this work.
The memory malloc
gives you is typically not executable. Jumping there causes a bus error (SIGBUS
). Assuming you are on a POSIX-like system, either allocate the memory for the function with mmap
and flags that cause the memory region to be executable or use mprotect
to mark the region as executable.
You also need to be more careful with the amount of memory you provide, you cannot simply take the size of a function and expect that to be the length of the function, sizeof
is not designed to provide this kind of functionality. You need to find out the function length using some other approach.
On modern desktops, the virtual memory manager is going to get in your way. Memory regions have three types of access: read, write, and execute. On systems where code segments have only execute permission, the memcpy
will fail with a bus error. In the more typical case, where only code segments have the execute permission, you can copy the function, but not run, because the memory region that contains bar
will not have execute permission.
Also, determining the size of the function is problematic. Consider the following program
void foo( int *x )
{
printf( "x:(%zu %zu) ", sizeof x, sizeof *x );
}
int main( void )
{
int x = 0;
foo( &x );
printf( "foo:(%zu %zu)\n", sizeof foo, sizeof *foo );
}
On my system, the output is x:(8 4) foo:(1 1)
indicating that taking the sizeof
a function pointer, or the function itself, is not a supported operation.
来源:https://stackoverflow.com/questions/33964277/dereferencing-function-pointers-in-c-to-access-code-memory