Thread: C memory allocation to c++

Hi,

I would like to convert my c style memory allocation (calloc) to c++ style (new and delete). The way I am doing it at the moment in c is as follows:

Code:

char** seq_array;
seq_array = (char **)calloc( (nseqs) * sizeof (char *), sizeof(char) );

for(int i=0; i < nseqs; i++)
seq_array[i] = (char *)calloc((length) * sizeof (char), sizeof(char));

This is to generate an array of size nseqs*length. I would like to change this to the c++ way of allocating memory. Would it be something like the following:

Code:

seq_array = char**[nseqs][length];

And what about freeing up the memory?

Thanks for your help
Mark

Code:

char **seq_array;
seq_array = new (char *)[nseqs];
for (int i = 0; i < nseqs; ++i)
  seq_array = new char[length];

The only differences is that, unlike calloc(), operator new does not guarantee initialisation to zero for basic types. To achieve that, modify the above to;

Code:

char **seq_array;
seq_array = new (char *)[nseqs];
for (int i = 0; i < nseqs; ++i)
{
    seq_array = new char[length];
    for (int j = 0; j < length; ++j)
       seq_array[i][j] = 0;
}

Incidentally, the original versions with calloc() contains a minor error. It should be

Code:

char** seq_array;
seq_array = (char **)calloc( (nseqs) * sizeof (char *), sizeof(char *) );

for(int i=0; i < nseqs; i++)
seq_array[i] = (char *)calloc((length) * sizeof (char), sizeof(char));

In this example, it doesn't matter as the subsequent loop overwrites each element of seq_array, but remove that loop ......

Thanks very much for you reply