需要从3d阵列到1d阵列的memcpy解释

Question

Why can't I use memcpy to flatten 3d dynamic allocated array to 1d array?

What is the difference between storage of 3d and 1d dynamic allocated array?

Thanks for your attention.I would be really appreciate if you could give some explanation/links/books.

#include<iostream>
using namespace std;
int main()
{               
        float ***b;
        int i,j,k;
        int ntab=26,ncrss=3,nsubs=1;
        b=(float***)malloc(ntab*ncrss*nsubs*sizeof(float));
        for(i=0;i<nsubs;i++)
        {
                b[i]=(float**)malloc(ncrss*ntab*sizeof(float));
                for(j=0;j<ncrss;j++)
                {
                        b[i][j]=(float*)malloc(ntab*sizeof(float));
                }
        }
        for(i=0;i<nsubs;i++)
                for(j=0;j<ncrss;j++)
                        for(k=0;k<ntab;k++)
                                b[i][j][k]=k;
        for(i=0;i<nsubs;i++)
                for(j=0;j<ncrss;j++)
                        for(k=0;k<ntab;k++)
                                printf("b[%d][%d][%d]=%f\n",i,j,k,b[i][j][k]);
        float *a;
        a=(float*)malloc(ntab*ncrss*nsubs*sizeof(float));
        //memcpy(a,b,ntab*ncrss*nsubs*sizeof(float));
        for(i=0;i<nsubs;i++)
        {
                const int index1=i*ncrss*ntab;
                for(j=0;j<ncrss;j++)
                {
                        const int index2=j*ntab+index1;
                        for(k=0;k<ntab;k++)
                        {
                                a[index2]=b[i][j][k];
                        }
                }
        }
        for(i=0;i<ntab*ncrss*nsubs;i++)
                printf("a[%d]=%f\n",i,a[i]);

        return 0;
}

Answer 1

You have multiple calls to malloc, which means your 3d array is not contiguous in memory. Thus memcpy cannot handle it all at once.

If you had allocated your 3d array in a contiguous fashion:

malloc(sizeof(float) * ntab * ncrss * nsubs)

then memcpy would work.

Answer 2

The 3d dynamic array which was constructed in the code snippet is a pointer to an array of pointers to arrays of pointers to arrays of values.

When memcpy is used with b as the source, it means the array to which b points to is copied. It means the array of pointers to arrays of pointers to arrays of values is copied.

In this case, the line

//memcpy(a,b,ntab*ncrss*nsubs*sizeof(float));

copies ntab*ncrss*nsubs*sizeof(float) bytes starting from the address to which b points to.