How to assign some part of dynamic array to whole static array

The size of dynamic array is the twice the size of stat开发者_JAVA技巧ic array. I want to assign the values which starts from (N/2)-1 to N-1 of dynamic array to whole static array.

The only way is copying the values with a loop?

My code:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main(int argc, char *argv[])
{
    int N=100, pSize=4, lSize, i;
    double *A;

    lSize=N/sqrt(pSize);
    /* memory allocation */
    A=(double*)malloc(sizeof(double)*N);

    double B[lSize];
    /* memory allocation has been done */

    /* initilize arrays */
    for(i=0; i<lSize; i++){
        B[i]=rand()% 10;
    }

    A=B;
    for (i=0; i<lSize; i++){
        fprintf(stdout,"%f\n", A[i]);
    }


    return 0;
}

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You can use the memcpy function to copy the data. For your example you want to copy the last half of A to B so could do something like:

memcpy(&B[0], &A[lSize-1], lSize * sizeof(double));

Note: On the MinGW compiler I was using, it was requiring that I declare the destination as &B[0], I thought I could get away with just B. It may be due to configuration I have (I don't use the C compiler all that much, normally just use g++ for quick C++ test cases).

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You can use memcpy to copy contiguous chunks of memory around.

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Your program leaks your allocation, which is probably bad - is that A=B intended to be where you would put the code that copies the array?

It may be possible, depending on your architecture, to do a copy without a CPU loop (via a call to a DMA engine or something). In standard C, you have no choice but to loop. You can either do it yourself or you can call memcpy(3), memmove(3), or bcopy(3) if you prefer to use the library's implementations.

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As said, you need to use memcpy:

#define  N  100
int staticarray[N];
int *pointer = (int*) malloc( sizeof(int)*N*2 );
memcpy( staticarray, (pointer + ((N/2) - 1)), sizeof(int)*N );