将数组与整数打印为十六进制

Question

I am using an API that uses a struct to represent an array (and allows filling up that array while accessing the struct).

If data is a struct object, and direction is a uint32_t, run the following:

printf("0x%08X", data->magic);

I get the value: 0xAAAABEEF

while printing the array directly as such:

printf("0x");
for (int i = 0; i < size; ++i) {
  printf("%02X", payload[i]);
}

I get the value: 0xEFBEAAAA

the struct definition goes like this:

struct Data {
  uint32_t magic;
} __attribute__((packed));

I believe the data variable is declared something like this:

// Declared and initialized somewhat like this:
uint8_t payload[kMaxSize];
Data* data = reinterpret_cast<Data*>(payload);
data->magic = 0xAAAABEEF;

I am curious why the printf does not return the same value. Is it because the machine is storing the data as LSB (least significant byte)?

Answer 1

Your guess is correct. In a little-endian processor (eg: x86 ), the least significant byte is stored first in memory. So the number 0xAAAABEEF will be stored as four bytes in memory: {0xEF, 0xBE, 0xAA, 0xAA}

When your program looks at those four bytes in memory, the way the data is interpreted - its type - determines how it looks. If {0xEF, 0xBE, 0xAA, 0xAA} is interpreted as individual bytes, you get "EF BE AA AA". But if it's interpreted as a uint32_t , then the computer knows to reverse the order and display it as "0xAAAABEEF".