To Write a C Program to take 32 bit integer and display Higher and lower words (16-bits) using union-CodePudding

union breakbit {
    int32_t data;
    int16_t low;
    int16_t high;
} var;

int main() {
    var.data = 0x12345678;
    printf("Initial value:%x\n",var.data);
    printf("Higher bit value:%x\n",var.uplow.high);
    printf("Higher bit value:%x\n",var.uplow.low);

    return 0;
}

The output of the above code is only lower bits and not higher, so anyone can help finding the higher bits value?

CodePudding user response：

In your example, all three data, low and high overlay one-another.

As per your printf statement, declare the union so:

union breakbit {
    uint32_t data;
    struct {
        uint16_t low;
        uint16_t high;
    } uplow;
} var;

The meaning of 'low' and 'high` will be different on different machines. (See "Big Endian/Little Endian")

EDIT: When dealing with hex values, you probably want to use an unsigned datatype. I've altered this example to conform.

CodePudding user response：

To addition to @Fe2O3 good answer and untangle the endian:

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

union breakbit {
  uint32_t data32;
  uint16_t data16[2];
};

// Using a C99 compound literal to index the LS half.
#define LS_INDEX ((const union breakbit){ .data32 = 1}.data16[1])

int main(void) {
  union breakbit var = {.data32 = rand() * (RAND_MAX   1ul)   rand()};
  printf("Initial value:lx\n", (unsigned long) var.data32);

  printf("Higher bits value:x\n", var.data16[!LS_INDEX]);
  printf(" Lower bits value:x\n", var.data16[LS_INDEX]);
}

Output:

Initial value:c0b18ccf
Higher bits value:c0b1
 Lower bits value:8ccf