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?
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.
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:%08lx\n", (unsigned long) var.data32);
printf("Higher bits value:%04x\n", var.data16[!LS_INDEX]);
printf(" Lower bits value:%04x\n", var.data16[LS_INDEX]);
}
Output:
Initial value:c0b18ccf
Higher bits value:c0b1
Lower bits value:8ccf
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