Splitting char array into chunks quickly c++ - c++

Question

Is there a faster/ more efficient way to split a char array into chunks (say 21 chars/ array) other than looping over the whole array?
This is my attempt right now
const char* arr = line.c_str();
char smallArr[(int)ceil((double)strlen(arr)/(double)21)][21];
int arrSisze[(int)ceil((double)strlen(arr)/(double)21)][1];
int m=0;int n=0;
for (int i=0; i<strlen(arr); i++) {
smallArr[m][i]=arr[i];
arrSisze[m][0]=(i+1)-n;
if ((i-n)==19) {
m++;
n+=20;
}
}

Answer 1

1) Using memcpy
char myname[] = "hello";
char dest[20] = {0};
/* using memcpy to copy string: */
memcpy ( dest, myname, 5);
2) using strncpy
char str1[]= "To be or not to be";
char str2[40];
strncpy ( str2, str1, sizeof(str2) );

Answer 2

Yes, pointer arithmetic and memcpy. However, seeing as you're using C++, let's stick to std::strings and assume they do a memcpy for us.
std::vector<std::string> output;
output.reserve(line.length() / 21 + (line.length() % 21) ? 1 : 0);
auto i = line.begin(), j = i + 21;
for(; line.end() - j > 21; i = j, j+= 21)
{
output.emplace(i, j)
}
if(j != line.end())
{
output.emplace(j, line.end());
}
So, what is going on here? For our purpose, it is good enough to think of a std::string as a char array and a length variable.
First, we reserve enough space for our output. You did this too.
Next, we define 2 variables i and j. i represents the beginning of the current substring and j the one-past-the-end iterator. Iterators here can be thought of as pointers into the internal structure of the string's char array - they may even be char*s!
We then iterate over the original string a line at a time. emplace just constructs a new element of a vector in the correct place. The call is equivalent to output.push_back(std::string(i, j)) and is only available in C++11.
To finish, we check whether there is another whole chunk using line.end() - j > 21; line.end() is the one-past-the-end iterator for the char array - it points at the NUL character (if there is one). If there is not a whole chunk, we check for a partial chunk using j != line.end().

Answer 3

Don't re-optimize the standard library.
If you have a std::string, just use it.
Don't use floating point for integer calculations: just use integer arthmetic porperly
the problem complexity is O(1). No other solution will cover it with less of one string walk span, and related copies
Use C++ properly, and forget C.
=
std::vector<std::string> chunks;
chunks.resize(21);
size_t chunksize = line.size()/21+1;
for(size_t i=0,j=0; i<line.size(); i+=chunksize, ++j)
{ chunks[j] = line.substr(i,chunksize); }
Note that sting::size takes N^0, while strlen takes N^1 complexity (it has a loop inside).
In this code, my loop is 21^1 on chunks and substr is (N/21)^1 each chunk content, giving N^1 on the whole length.
There is no need to track string length and null terminate the strings. All is handled by std::string class.