How to copy memory

Question

Say i have:

unsigned char *varA, *varB, *varC;
varA=malloc(64);
varB=malloc(32);
varC=malloc(32);

How can i put the first 32 byte of varA into varB and the last 32 byte of varA into varC?

Answer 1

memcpy(varB, varA, 32);
memcpy(varC, varA + 32, 32);

It's this simple because the underlying data type is unsigned char, which is the same size as a byte. If varA, varB, and varC were integers, you would need to multiply the size parameter to memcpy (i.e. 32) by sizeof(int) to compute the right number of bytes to copy. If I were being pedantic, i could have multiplied 32 by sizeof(unsigned char) in the example above, but it is not necessary because sizeof(unsigned char) == 1.

Note that I don't need to multiply the 32 in varA + 32 by anything because the compiler does that for me when adding constant offsets to pointers.

One more thing: if you want to be fast, it might be sufficient to just work on each half of varA separately, rather than allocate two new buffers and copy into them.

Answer 2

You could use loop to copy individual bytes one by one:

for (int i = 0; i != 32; ++i)
    varB[i] = varA[i];

for (int i = 0; i != 32; ++i)
    varC[i] = varA[32 + i];

Or memcpy function from the C runtime library:

memcpy(varB, varA, 32);
memcpy(varC, varA + 32, 32);