Sometimes we need a version of memset which sets a value that is larger than one byte. memset only uses one byte of the value passed in and does byte-wise initialization. If you want to initialize an array with a particular value larger than one byte, just use these approaches:
1. Just use a simple for ( ; ; ) In most cases this is a simple and good way.
2. std::fill, std::fill_n
#define BUFF_SIZE 1024 * 1024 * 1536
...
uint16_t val16 = 10000;
uint32_t val32 = 10000 * 10000L;
uint64_t val64 = 10000 * 10000 * 10000LL;
void* buff = new uint8_t[BUFF_SIZE];
...
std::fill_n((uint16_t*)buff, BUFF_SIZE / sizeof(val16), val16);
std::fill_n((uint32_t*)buff, BUFF_SIZE / sizeof(val32), val32);
std::fill_n((uint64_t*)buff, BUFF_SIZE / sizeof(val64), val64);
3. The memset function is very optimized in the compilers. Some implementations offer a 16-bit version memsetw, but that's not everywhere. The memcpy implementations are often written with SIMD instructions which makes it possible to shuffle 128 bits at a time. SIMD instructions are assembly instructions that can perform the same operation on each element in a vector up to 16 bytes long. That includes load and store instructions. So, here is a memsetx function which is implemented via memcpy:
template <typename T>
void memsetx(void* dst, T& val, unsigned int size)
{
uint32_t i = 0;
for ( ; i < (size & (~(sizeof(T) - 1))); i += sizeof(T))
memcpy((uint8_t*)dst + i, &val, sizeof(T));
for ( ; i < size; ++i)
((uint8_t*)dst)[i] = ((uint8_t*)&val)[i & (sizeof(T) - 1)];
}
...
memsetx(buff, val16, BUFF_SIZE);
memsetx(buff, val32, BUFF_SIZE);
memsetx(buff, val64, BUFF_SIZE);
...
memsetx(buff, val1024, BUFF_SIZE);
...
