How to memset() memory to a certain pattern instead of a single byte?

Question

I need to write a repeating pattern to memory (e.g. 0x11223344), so that the whole memory looks like (in hex):

1122334411223344112233441122334411223344112233441122334411223344...

I can't figure out how to do it with memset() because it takes only a single byte, not 4 bytes.

Any ideas?

Answer 1

On OS X, one uses memset_pattern4( ) for this; I would expect other platforms to have similar APIs.

I don't know of a simple portable solution, other than just filling in the buffer with a loop (which is pretty darn simple).

Answer 2

An efficient way would be to cast the pointer to a pointer of the needed size in bytes (e.g. uint32_t for 4 bytes) and fill with integers. It's a little ugly though.

char buf[256] = { 0, }; uint32_t * p = (uint32_t *) buf, i;  for (i = 0; i < sizeof(buf) / sizeof(* p); i++) {     p[i] = 0x11223344; } 

Not tested!

Answer 3

Recursively copy the memory, using the area which you already filled as a template per iteration O(log(N)):

int fillLen = ...;
int blockSize = 4; // Size of your pattern

memmove(dest, srcPattern, blockSize);
char * start = dest;
char * current = dest + blockSize;
char * end = start + fillLen;
while(current + blockSize < end) {
    memmove(current, start, blockSize);
    current += blockSize;
    blockSize *= 2;
}
// fill the rest
memmove(current, start, (int)end-current);

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What I mean with O(log(N)) is that the runtime will be much faster than if you fill the memory manually since memmove() usually uses special, hand-optimized assembler loops that are blazing fast.