c++ pass char array address in 64bit platform

Question

i have func() which must have to take unsigned int parameter.

there is a reason for void func(unsigned int val),

i have to pass various type(unsigned char, unsigned short, unsigned int) to func parameter.

and i have to pass char array to func(), my current solution like below code.

Edit: is there easy way to port this code in 64bit platform?

char test_str[128] = { 0 };
void func(unsigned int val)
{
  memcpy(test_str, (char *)val, 128); //current my solution
  printf("%s\n", test_str);
}

int main()
{
  char str[128] = "hello world";
  func((unsigned int)(char *)&str); //current my solution
  return 0;
}

note: intptr_t

Is this C++? When I see lots of casts it usually indicates the programmer is doing something wrong — Ed Heal, Mar 24 '17 at 02:40
function must have unsigned int, should i change it to template? — sailfish009, Mar 24 '17 at 02:42
Why should it have an `unsigned int`? Why using `printf` for C++ code? Why have you tagged it C? — Ed Heal, Mar 24 '17 at 02:43
This sounds like [an XY problem](https://meta.stackexchange.com/questions/66377/what-is-the-xy-problem). — R Sahu, Mar 24 '17 at 02:45
They are two different languages - with different mechanisms to solve the use case (whatever this is) — Ed Heal, Mar 24 '17 at 02:48
Pass the address of a char array as follows: `DoSomething(&charArray);` Apart from that, I have no idea what you're asking? — Disillusioned, Mar 24 '17 at 02:58

score 4 · Accepted Answer · answered Mar 24 '17 at 03:07

There are a lot of problem with your code, but at least for me the one the compiler first complains about is:

error: cast from pointer to smaller type 'unsigned int' loses information │Offset: 4 byte: 0x7ffe4b93ddd4 contents:9 func((unsigned int)(char *)str);

I assume that you're trying to sneak in the literal address of the char array into the unsigned int parameter. However an unsigned int can only hold 4 bytes (in my platform), that is not enough to hold the full address, since all pointers require 8 bytes (again, in my platform). See for yourself.

#include <stdio.h>

int main (int argc, char  *argv[])
{
    char str[128] = "hello world";
    unsigned int *address = (unsigned int *)str;

    printf("address: %p \t contains: %s \t pointer size: %lu \n", address, (char *)address, sizeof(char *));
    // address: 0x7ffcf9492540          contains: hello world          pointer size: 8
    printf("Size of address in pointer: %lu \n", sizeof(long long *));
    // will print the same size as the past sizeof() operation
    printf("Size of unsigned int variable: %lu \n", sizeof(unsigned int));
    // 4 bytes, not enough to fit in the necessary 8 bytes

    return 0;
}

https://msdn.microsoft.com/en-us/library/aa384267(VS.85).aspx found some helper functions — sailfish009, Mar 24 '17 at 04:08

c++ pass char array address in 64bit platform

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