Why take the address of an array element like `&myArray[i]`? Bidimensional array

Question

There's some code where it creates a float array like this:

mData = new float[channelCount * maxFrames];

then it does

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

What does &mData[sampleIndex] mean? Well, we have a float array, we take an element of that array, and then take the address of that element. Wouldn't the address of that element be mData + sampleIndex?

What if I wanted to change memcpy by a for loop? I did this and it worked:

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

but I don't know what (&mData[sampleIndex])[i] means. Should it be mData + sampleIndex + i?

This code is supposed to work to record microfone wav data, so we should be able to store things in multiple channels. How this code manages such channels?

Your loop is wrong, it should be for (int i=0; i< numSamples; i++) since you are assigning whole floats and not individual bytes, assuming buffer[] is an array of float to match that mData is an array of float. — Remy Lebeau
– Remy Lebeau, Commented Sep 15, 2020 at 0:04
memcpy(&mData[sampleIndex], buffer, (numSamples * sizeof(float))); is useful when you want to memcpy data into the middle of an array. — user4581301
– user4581301, Commented Sep 15, 2020 at 0:04

eerorika · Accepted Answer · 2020-09-15 00:15:10Z

2

What does &mData[sampleIndex] mean? Well, we have a float array, we take an element of that array, and then take the address of that element. Wouldn't the address of that element be mData + sampleIndex?

Yes.

What if I wanted to change memcpy by a for loop?

Your loop doesn't quite do the same thing. memcpy is copying numSamples * sizeof(float) bytes while your loop is copying numSamples * sizeof(float) floats. Since a float consists of multiple bytes (on most systems), this may result in a buffer overflow.

but I don't know what (&mData[sampleIndex])[i] means. Should it be mData + sampleIndex + i?

It's not quite the same. (&mData[sampleIndex])[i] would be equal to *(mData + sampleIndex + i)

How this code manages such channels?

This code simply copies values from one array into another. It doesn't "manage" anything.

answered Sep 15, 2020 at 0:15

eerorika

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2 Comments

Poperton Over a year ago

How can I copy numSamples * sizeof(float) bytes instead of floats?

eerorika Over a year ago

@GuerlandoOCs by copying numSamples floats.

Remy Lebeau · Accepted Answer · 2020-09-15 00:12:39Z

1

The syntax array[index] is the same as *(array + index), thus:

&mData[sampleIndex]

is the same as:

&(*(mData + sampleIndex))

Which is simply:

mData + sampleIndex

And so, (&mData[sampleIndex])[i] is getting a float* pointer to the mData element at index sampleIndex, and then applying the index i to that pointer. So yes, in this case:

(&mData[sampleIndex])[i] = buffer[i];

is the same as:

*(mData + sampleIndex + i) = *(buffer + i);

edited Sep 15, 2020 at 0:12

answered Sep 15, 2020 at 0:06

Remy Lebeau

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