Pass uint8_t* as parameter to raw function pointer

Question

I have the problem where I want to pass a uint8_t[] array as a parameter to a function pointer defined as `typedef void (dangerousC)(void); Also, I'm using Windows API headers.

Assume the variable raw is a function pointer returned by GetProcAddress(). Also assume that the parameters to foo() are not known by the compiler.

Here is the full code:

#include <iostream>
#include <Windows.h>

typedef void (*dangerousC)(char*);

void foo(int a) {
   std::cout << a << std::endl;
}

int main() {
    auto raw = (FARPROC) foo;

    auto* b = new uint8_t[4];
    b[0] = 74;
    b[1] = 35;
    b[2] = 0;
    b[3] = 0;

    std::cout << *(int*)b << std::endl;
    auto func = (dangerousC)raw;
    func(reinterpret_cast<char *>(*b));
    delete[] b;
}

When I call the function pointer with the parameters reinterpret_cast<char *>(*b), I only get one character, which is reasonable when I dereference a pointer.

But I also tried with a pointer and it does not print the result that I want, which is 9034.

How could I make the function print 9034 and fully interpret the byte array as a 32-bit int?

Answer 1

foo() expects an int, but you are passing it a char* instead. As such, you need to pass in a char* whose value is the integer you want, not an actual pointer to an integer.

#include <iostream>
#include <Windows.h>

typedef void (*dangerousC)(char*);

void foo(int a) {
   std::cout << a << std::endl;
}

int main() {
    auto raw = (FARPROC) foo;

    uint8_t b[4];
    b[0] = 74;
    b[1] = 35;
    b[2] = 0;
    b[3] = 0;

    int32_t i = *reinterpret_cast<int32_t*>(b);
    std::cout << i << std::endl;
    auto func = reinterpret_cast<dangerousC>(raw);
    func(reinterpret_cast<char*>(i));
}

Online Demo

Answer 2

Usually it's done like below, as I understood it from your comment properly, you deal with byte streams

int n;
std::memcpy(&n, b, sizeof(n));
func(b);