为什么将“extern puts”转换为函数指针“（void（*）（char *））＆puts”？

Question

I'm looking at example abo3.c from Insecure Programming and I'm not grokking the casting in the example below. Could someone enlighten me?

int main(int argv,char **argc)   
{  
    extern system,puts;  
    void (*fn)(char*)=(void(*)(char*))&system;  
    char buf[256];  

    fn=(void(*)(char*))&puts;  
    strcpy(buf,argc[1]);  
    fn(argc[2]);  
    exit(1);  
}

So - what's with the casting for system and puts? They both return an int so why cast it to void?

I'd really appreciate an explanation of the whole program to put it in perspective.

[EDIT]
Thank you both for your input!

Jonathan Leffler , there is actually a reason for the code to be 'bad'. It's supposed to be exploitable, overflowing buffers and function pointers etc. mishou.org has a blog post on how to exploit the above code. A lot of it is still above my head.

bta , I gather from the above blog post that casting system would somehow prevent the linker from removing it.

One thing that is not immediately clear is that the system and puts addresses are both written to the same location, I think that might be what gera is talking about “so the linker doesn't remove it”.

While we are on the subject of function pointers, I'd like to ask a follow-up question now that the syntax is clearer. I was looking at some more advanced examples using function pointers and stumbled upon this abomination, taken from a site hosting shellcode.

#include <stdio.h>

char shellcode[] = "some shellcode";

int main(void)
{
    fprintf(stdout,"Length: %d\n",strlen(shellcode));
    (*(void(*)()) shellcode)();
}

So the array is getting cast to a function returning void , referenced and called? That just looks nasty - so what's the purpose of the above code?

[/EDIT]

Answer 1

Original question

User bta has given a correct explanation of the cast - and commented on the infelicity of casting system .

I'm going to add:

The extern line is at best weird. It is erroneous under strict C99 because there is no type, which makes it invalid; under C89, the type will be assumed to be int . The line says 'there is an externally defined integer called system, and another called puts', which is not correct - there are a pair of functions with those names. The code may actually 'work' because the linker might associate the functions with the supposed integers. But it is not safe for a 64-bit machine where pointers are not the same size as int . Of course, the code should include the correct headers ( <stdio.h> for puts() and <stdlib.h> for system() and exit() , and <string.h> for strcpy() ).

The exit(1); is bad on two separate counts.

• It indicates failure - unconditionally. You exit with 0 or EXIT_SUCCESS to indicate success.

• In my view, it is better to use return at the end of main() than exit() . Not everyone necessarily agrees with me, but I do not like to see exit() as the last line of main() . About the only excuse for it is to avoid problems from other bad practices, such as functions registered with atexit() that depend on the continued existence of local variables defined in main() .

---

/usr/bin/gcc -g -std=c99 -Wall -Wextra -Wmissing-prototypes -Wstrict-prototypes -Wold-style-definition -c nasty.c
nasty.c: In function ‘main’:
nasty.c:3: warning: type defaults to ‘int’ in declaration of ‘system’
nasty.c:3: warning: type defaults to ‘int’ in declaration of ‘puts’
nasty.c:3: warning: built-in function ‘puts’ declared as non-function
nasty.c:8: warning: implicit declaration of function ‘strcpy’
nasty.c:8: warning: incompatible implicit declaration of built-in function ‘strcpy’
nasty.c:10: warning: implicit declaration of function ‘exit’
nasty.c:10: warning: incompatible implicit declaration of built-in function ‘exit’
nasty.c: At top level:
nasty.c:1: warning: unused parameter ‘argv’

Not good code! I worry about a source of information that contains such code and doesn't explain all the awfulness (because the only excuse for showing such messy code is to dissect it and correct it).

---

There's another weirdness in the code:

int main(int argv,char **argc)   

That is 'correct' (it will work) but 100% aconventional. The normal declaration is:

int main(int argc, char **argv)

The names are short for 'argument count' and 'argument vector', and using argc as the name for the vector (array) of strings is abnormal and downright confusing.

---

Having visited the site referenced, you can see that it is going through a set of graduated examples. I'm not sure whether the author simply has a blind spot on the argc/argv issue or is deliberately messing around ( 'abo1' suggests that he is playing, but it is not helpful in my view). The examples are supposed to feed your mind, but there isn't much explanation of what they do. I don't think I could recommend the site.

---

Extension question

What's the cast in this code doing?

#include <stdio.h>

char shellcode[] = "some shellcode";

int main(void)
{
    fprintf(stdout,"Length: %d\n",strlen(shellcode));
    (*(void(*)()) shellcode)();
}

This takes the address of the string 'shellcode' and treats it as a pointer to a function that takes an indeterminate set of arguments and returns no values and executes it with no arguments. The string contains the binary assembler code for some exploit - usually running the shell - and the objective of the intruder is to get a root-privileged program to execute their shellcode and give them a command prompt, with root privileges. From there, the system is theirs to own. For practicing, the first step is to get a non-root program to execute the shellcode, of course.

Reviewing the analysis

The analysis at Mishou 's web site is not as authoritative as I'd like:

One, this code uses the extern keyword in the C language to make the system and puts functions available. What this does (I think) is basically references directly the location of a function defined in the (implied) header files…I get the impression that GDB is auto-magically including the header files stdlib.h for system and stdio.h for puts. One thing that is not immediately clear is that the system and puts addresses are both written to the same location, I think that might be what gera is talking about “so the linker doesn't remove it”.

Dissecting the commentary:

1. The first sentence isn't very accurate; it tells the compiler that the symbols system and puts are defined (as integers) somewhere else. When the code is linked, the address of puts() -the-function is known; the code will treat it as an integer variable, but the address of the integer variable is, in fact, the address of the function - so the cast forces the compiler to treat it as a function pointer after all.
2. The second sentence is not fully accurate; the linker resolves the addresses of the external 'variables' via the function symbols system() and puts() in the C library.
3. GDB has nothing whatsoever to do the compilation or linking process.
4. The last sentence does not make any sense at all. The addresses only get written to the same location because you have an initialization and an assignment to the same variable.

This didn't motivate me to read the whole article, it must be said. Due diligence forces me onwards; the explanation afterwards is better, though still not as clear as I think it could be. But the operation of overflowing the buffer with an overlong but carefully crafted argument string is the core of the operation. The code mentions both puts() and system() so that when run in non-exploit mode, the puts() function is a known symbol (otherwise, you'd have to use dlopen() to find its address), and so that when run in exploit mode, the code has the symbol system() available for direct use. Unused external references are not made available in the executable - a good thing when you realize how many symbols there are in a typical system header compared with the number used by a program that includes the header.

There are some neat tricks shown - though the implementation of those tricks is not shown on the specific page; I assume (without having verified it) that the information for getenvaddr program is available.

The abo3.c code can be written as:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(int argc, char **argv)   
{  
    void (*fn)(char*) = (void(*)(char*))system;  
    char buf[256];  

    fn = (void(*)(char*))puts;  
    strcpy(buf, argv[1]);  
    fn(argv[2]);  
    exit(1);  
}

Now it compiles with only one warning with the fussy compilation options I originally used - and that's the accurate warning that 'argc' is not used. It is just as exploitable as the original; it is 'better' code though because it compiles cleanly. The indirections were unnecessary mystique, not a crucial part of making the code exploitable.

Answer 2

Both system and puts normally return int . The code is casting them to a pointer that returns void , presumably because they want to ignore whatever value is returned. This should be equivalent to using (void)fn(argc[2]); as the penultimate line if the cast didn't change the return type. Casting away the return type is sometimes done for callback functions, and this code snippet seems to be a simplistic example of a callback.

Why the cast for system if it is never used is beyond me. I'm assuming that there's more code that isn't shown here.