How to transform this memcpy into a for?

Question

What is the difference between these two?

---

for (int i = 0; i < numSamples; i++) {
    mData[sampleIndex++] = *buffer++ * (1.0f / 32768);
}

and

memcpy(&mData[sampleIndex], buffer, (numSamples * sizeof(float)));

---

If I understood correct, the first copies numSamples float values to mData, one by one. The second one, copies numSamples*sizeof(float) bytes to mData. Since we're copying numSaples * number of bytes on float, I think they do the same thing, but the first one actually multiplies things before passing to mData.

So, is there a way to transform the memcpy into a for? Something like:

for (int i = 0; i < numSamples * sizeof(float); i++) {
    //What to put here?
}

Context:

const int32_t   mChannelCount;
const int32_t   mMaxFrames;
int32_t sampleIndex;
float          *mData;
float *buffer;

Answer 1

I gather from your post that you want to make a memcpy similar copy but using a for loop, that being the case you just need do use the same for loop but without the multiplication part:

for (int i = 0; i < numSamples; i++){
    mData[sampleIndex++] = *buffer++;
}

Note that memcpy can be more effective than a for loop given the conditions (see Maxim Egorushkin and Jeremy Friesner comments bellow) so you may want to keep it that way.

Another, more idiomatic, and, I would argue, better way to implement the operations you are performing is to use the C++ library provided methods as sugested by Ted Lyngmo and rustyx.

Disclaimer: As I was writing my answer, Martin York posted a comment with a similar solution, that being the case, credit to him as well.

Answer 2

What is the difference between these two?

The former performs a calculation on the source array while copying the result into another array a float at a time.

The latter copies the content of the array byte at a time into another without calculation.

So, is there a way to transform the memcpy into a for?

Yes. Here is a naïve way to transform it:

auto dest_c = static_cast<unsigned char*>(mData + sampleIndex);
auto src_c = static_cast<const unsigned char*>(buffer);
auto end = src_c + numSamples * sizeof(float);
for (; src_c < end;) { // or while(src_c < end)
    *dest_c++ = *src_c++;
}

The actual implementation of the standard function is likely more complex, involving optimisations related to copying long sequences.

---

Since you don't appear to need the generic reinterpretation aspect of std::memcpy, perhaps a simpler alternative would suffice:

auto dest = mData + sampleIndex;
auto src = buffer;
auto end = src + numSamples;
for (; src < end;) {
    *dest++ = *src++;
}

Or perhaps another standard algorithm:

std::copy(buffer, buffer + numSamples, mData + sampleIndex);

Answer 3

What is the difference between these two?

for (int i = 0; i < numSamples; i++) {
  mData[sampleIndex++] = *buffer++ * (1.0f / 32768);
}
// and
memcpy(&mData[sampleIndex], buffer, (numSamples * sizeof(float)));

These are quite different given the * (1.0f / 32768);. I assume the code compare is setting the scaling difference aside. @Thomas Matthews.

• Important: buffer, sampleIndex has different values after the for loop.

• *buffer++ needs no code change should the type of buffer change. * sizeof(float) obilgies a code change. Could have used * sizeof *buffer.

• mempcy() is optimized code per that platform. for() loops can only do so much. In particular, mempcy() assumes mData, buffer do not overlap. The for() loop may not be able to make that optimization.

• This for uses int indexing where memcpy() uses size_t. Makes a difference with huge arrays.

• memcpy() tolerates an unaligned pointers. mData[sampleIndex++] = *buffer++ .. does not.

"the first copies numSamples float values to mData, one by one. " is a not certain. A smart compiler may be able to make certain parallel copies depending on the context and act as if copying was done one by one.

Post the entire block of code/function that uses these 2 approaches for a better compare.