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

user_257942 0
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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?

memset memory c design-patterns
7 Answers

Standard C library has no such function. But memset is usually implemented as an unrolled loop to minimize branching and condition checking:

static INLINE void memset4(uint32_t *RESTRICT p, uint32_t val, int len) {
  uint32_t *end = p + (len&~0x1f); //round down to nearest multiple of 32
  while (p != end) { //copy 32 times
    p[ 0] = val;
    p[ 1] = val;
    p[ 2] = val;
    p[ 3] = val;
    p[ 4] = val;
    p[ 5] = val;
    p[ 6] = val;
    p[ 7] = val;
    p[ 8] = val;
    p[ 9] = val;
    p[10] = val;
    p[11] = val;
    p[12] = val;
    p[13] = val;
    p[14] = val;
    p[15] = val;
    p[16] = val;
    p[17] = val;
    p[18] = val;
    p[19] = val;
    p[20] = val;
    p[21] = val;
    p[22] = val;
    p[23] = val;
    p[24] = val;
    p[25] = val;
    p[26] = val;
    p[27] = val;
    p[28] = val;
    p[29] = val;
    p[30] = val;
    p[31] = val;
    p += 32;
  }
  end += len&0x1f; //remained
  while (p != end) *p++ = val; //copy remaining bytes
}

Good compiler will likely use some CPU specific instructions to optimize it further (like i.e. use SSE 128-bit store), but even without optimizations, it should be as fast as a library memset, because such simple loops are memory access bound.

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