copying a short int to a char array

Question

I have a short integer variable called s_int that holds value = 2

unsighed short s_int = 2;

I want to copy this number to a char array to the first and second position of a char array.

Let's say we have char buffer[10]; . We want the two bytes of s_int to be copied at buffer[0] and buffer[1] .

How can I do it?

Answer 1

The usual way to do this would be with the bitwise operators to slice and dice it, a byte at a time:

b[0] = si & 0xff;
b[1] = (si >> 8) & 0xff;

though this should really be done into an unsigned char , not a plain char as they are signed on most systems.

Storing larger integers can be done in a similar way, or with a loop.

Answer 2

*((short*)buffer) = s_int;

但是viator emptor得到的字节顺序会随着字节顺序而变化。

Answer 3

By using pointers and casts.

unsigned short s_int = 2;
unsigned char buffer[sizeof(unsigned short)];

// 1.
unsigned char * p_int = (unsigned char *)&s_int;
buffer[0] = p_int[0];
buffer[1] = p_int[1];

// 2.
memcpy(buffer, (unsigned char *)&s_int, sizeof(unsigned short));

// 3.
std::copy((unsigned char *)&s_int,
          ((unsigned char *)&s_int) + sizeof(unsigned short),
          buffer);

// 4.
unsigned short * p_buffer = (unsigned short *)(buffer); // May have alignment issues
*p_buffer = s_int;

// 5.
union Not_To_Use
{
  unsigned short s_int;
  unsigned char  buffer[2];
};

union Not_To_Use converter;
converter.s_int = s_int;
buffer[0] = converter.buffer[0];
buffer[1] = converter.buffer[1];

Answer 4

I would memcpy it, something like

memcpy(buffer, &s_int, 2);

The endianness is preserved correctly so that if you cast buffer into unsigned short *, you can read the same value of s_int the right way. Other solution must be endian-aware or you could swap lsb and msb. And of course sizeof(short) must be 2.

Answer 5

If you don't want to make all that bitwise stuff you could do the following

char* where = (char*)malloc(10);
short int a = 25232;
where[0] = *((char*)(&a) + 0);
where[1] = *((char*)(&a) + 1);