如何写入缓冲区（空指针）？

Question

I want to write in 2 chars and a bit vector (uint64_t) to a file, but I first have to write them all to a buffer. Then the buffer will be written to the file. How should I write these 3 variables into a buffer (void pointer) so that all can be contained within one (void pointer) variable.

For example I want to write

char a = 'a';
char b = 'b';
uint64_t c = 0x0000111100001111;

into

void *buffer = malloc(sizeof(char)*2+sizeof(uint64_t));

Then write that into a file using

write(fd, buffer, sizeof(char)*2+sizeof(uint64_t));

Answer 1

This is the (almost*) completely safe way of doing it:

uint8_t *buffer = malloc(2 + sizeof(uint64_t));
buffer[0] = a;
buffer[1] = b;
memcpy(buffer + 2, &c, sizeof(c));

You might be tempted to do something like *(uint64_t *)(buffer + 2) = c; but that's not portable due to alignment restrictions.

Note that sizeof(char) == 1 , per definition in the C standard.

(*) I've assumed 8-bit char , which is nearly, but not entirely universal; on a platform with 16-bit char , use memcpy for a and b as well.

Answer 2

Well, you can allways put the data into a buffer like this:

void *buffer = malloc(size_of_buffer);
char *ch = buffer;
*pos++ = a;
*pos++ = b;
*(uint64_t*)(pos) = c

Or use memcpy like cnicutar suggests. However, I would think it's a lot easier and less error prone to just write each element to the file one at the time:

write(fd, &a, sizeof a);
write(fd, &b, sizeof b);
write(fd, &c, sizeof c);

Beware of endian issues.