如何使用 integer 索引获得 64 位无符号长的 16 位部分？

Question

I'm trying to return certain 16-bit sections from a 64-bit unsigned long and I'm stuck on how to accomplish this. Here is the function I am trying to implement:

// assume i is a valid index (0-3 inclusive)
unsigned short get(unsigned long* ex, int i) {
  // return the 16-bit section based on the index i
}

For example, if I have unsigned long ex = 0xFEDCBA9876543210; , then my function get(ex, 0) would return 0x3210 , get(ex, 1) would return 0x7654 , etc. I'm very new to C and I'm still trying to wrap my head around bit management and pointers. Any advice or feedback is appreciated in helping me understand C better.

Answer 1

You'll have to use a bit shift

#include <stdint.h>
uint16_t get(uint64_t ex, int i)
{
    return (uint16_t)(ex >> i*16);
}

You can just pass by value. There's no reason to pass a pointer. This will shift the bits to the right, meaning they become the low order value. When it gets converted to a 16 bit type, it loses the higher order bits.

I've included stdint.h because it defines types of exact size.

Answer 2

I would use a mask along with bit shifting

unsigned short get(int index, unsigned long n)
{
    if (index > 3 || index < 0)
        return 0xFFFF; // well you have to see if you have control over the inputs.
    return (n >> (index << 4)) & 0xFFFF; // will extract 2 bytes.
}

Answer 3

The way closest to what you're asking for is a union:

#include <stdio.h>
#include <stdint.h>

int main(const int argc, const char * const argv[]) {
    union {
        uint64_t i;
        uint16_t ia[4];
    } u;
    u.i = 0xFEDCBA9876543210;
    printf("u.ia[0] %x\n", u.ia[0]);
}

Output is:

u.ia[0] 3210