Question

I am having problems with a piece of code. I have this line in a C program:

mel[i] = malloc(sizeof(double)*mellength[i]);

Where:

double *mel[16];
int mellength[16];

And I need to do this in C++ however I am using the “new” operator:

mel[i] = new double(sizeof(double)*mellength[i]);

Where:

int *mellength = new int[16];
double *mel = new double[16];

And I get the following error:

error: cannot convert ‘double*’ to ‘double’ in assignment

The full function:

void setup_Mel(int fft_size, int sample_rate, int *melstart, int *mellength, double *mel)
{

double fmax;
double dphi;
double fsample;
double freq;
double temp[fft_size/2];

fmax=2595*log10(8000.0f/700+1);
dphi = fmax/17;
freq = (double)sample_rate/fft_size;

for(int i=0; (i < 13); i++)
{
    melstart[i] = fft_size/2;
    mellength[i] = 0;

    memset(temp, 0, sizeof(double)*fft_size/2);
    for(int j=0; (j < fft_size/2); j++)
    {
        fsample = 2595*log10(freq*j/700 + 1);

        if((dphi * i <= fsample) && (fsample < dphi*(i+1)))
        {
            temp[j] = (fsample-dphi*i)/(dphi*(i+1)-dphi*i);
        }

        if ((dphi*(i+1) <= fsample) && (fsample < dphi*(i+2))) 
        {
            temp[j] = (fsample-dphi*(i+2))/(dphi*(i+1)-dphi*(i+2));
        }

        if ((temp[j] != 0) && (melstart[i] > j))
        {
            melstart[i] = j;
        }

        if (temp[j] != 0) mellength[i]++;
    }

   mel[i] = new double(sizeof(double)*mellength[i]);

}
}

Hope someone can help 🙂

Answer 1

mel[i] = new double(sizeof(double)*mellength[i]);

Doesn’t do what you think – it allocates a single double with the value sizeof(double)*mellength[i].

To allocate an array, you’d need

mel[i] = new double[mellength[i]];

Sample

The two versions are not equivalent:

double *mel = new double[16];

in C++ declares an array of double, whereas

double *mel[16];

declares an array of pointers to double. When you attempt to dereference the C++ variable, you get back a double, so of course you can’t assign it to the return of a new.