Calling a function via its address C


I tried to implement a solution of this question: Calling a function through its address in memory in c / c++, but I'm not very familiar with the differences in C and C++. When I try to implement the answer, my compiler throws a weird error message at me:

shellcode/findpattern.c: In function ‘shell_code’:
shellcode/findpattern.c:9:30: error: expected expression before ‘)’ token
     memchr* memchr = (memchr*)0xdeadbeef;
shellcode/findpattern.c:10:30: error: expected expression before ‘)’ token
     memcmp* memcmp = (memcmp*)0xdeadb00f;

Here is my code:

//#include "string.h"
#include "stdio.h"
//#include "stdlib.h"

    typedef void* memchr(const void* , int , size_t  );
    typedef int memcmp(const void* , const void* , size_t  );

void shell_code(){
    memchr* memchr = (memchr*)0xdeadbeef;
    memcmp* memcmp = (memcmp*)0xdeadb00f;

    unsigned char *current = 0x00400000;
    unsigned char *end = 0x015f1000;
    int patternlength = 8;
    unsigned char pattern[8] = "\x48\x08\x49\x8B\x48\x11\x8B\$
    unsigned char *ret;
    while(current < end){
        ret = memchr(current, pattern[0], end-current);
        if (ret != NULL){
            if (memcmp(current, &pattern, patternlength) == 0$
                return current + patternlength;
        current = ret;

What am I missing here? As far as I understand this is just a cast, so why does the compiler throw an error here? Is this a C vs C++ thing that I'm unfamiliar with?

asked on Stack Overflow Oct 21, 2018 by reijin

1 Answer


Given the code (extracted from your code):

typedef void* memchr(const void *, int, size_t);

void shell_code(void)
    memchr* memchr = (memchr*)0xdeadbeef;
//  ^1      ^2        ^3

The ^1 mention of memchr is to the function type in the typedef; the ^2 mention is the name of a local variable. This local variable now hides the type; you can no longer access the type in the function. The ^3 mention of memchr is a reference to the local pointer-to-function variable, not to the type. You have a multiplication operator after the variable, but no RHS for the multiplication — so the compiler complains about the ) because it expected an expression there as the RHS of the multiplication.

Don't play so hard with the same name. You'll confuse people reading your code. Use different names for the type and the function pointer. For example (not necessarily good naming style, but sufficient — and avoiding functions in favour of function pointers):

typedef void *(*MemChr)(const void *, int, size_t);

void shell_code(void)
    MemChr p_memchr = (MemChr)0xdeadbeef;

The code now stands a chance of compiling, but will simply crash when you run it and call p_memchr because the code at 0xDEADBEEF is unlikely to be a function like memchr, assuming it is mapped as executable code at all.

Note that this notation allows you to #include <string.h> (where memchr() is declared) without interfering with it, or interference from it.

answered on Stack Overflow Oct 21, 2018 by Jonathan Leffler • edited Oct 21, 2018 by Antti Haapala

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